【比赛链接】
Div. 1Div. 2【题解链接】
点击打开链接【Div.2 A】Mind the Gap
【思路要点】
从小到大枚举答案，检查合法性。时间复杂度\(O(Ans*N)\)。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 100005; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int n, s, a[MAXN]; bool valid(int x) { if (x + s < a[1]) return true; if (x - s > a[n]) return true; for (int i = 1; i < n; i++) if (x - s > a[i] && x + s < a[i + 1]) return true; return false; } int main() { read(n), read(s); for (int i = 1; i <= n; i++) { int x, y; read(x), read(y); a[i] = x * 60 + y; } int ans = 0; while (!valid(ans)) ans++; printf("%d %d\n", ans / 60, ans % 60); return 0; }【Div.2 B】Watering System
【思路要点】
显然我们会优先堵住大小较大的洞，排序后贪心即可。时间复杂度\(O(NLogN)\)。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 100005; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int main() { int n, A, B, C, T, ans = 0; read(n), read(A), read(B), read(C), read(T); for (int i = 1; i <= n; i++) { int x; read(x); ans += A; if (C >= B) ans += (T - x) * (C - B); } writeln(ans); return 0; }【Div.2 C/Div.1 A】Stairs and Elevators
【思路要点】
不难发现我们只有可能使用起点左起以及右起的第一个楼梯与电梯，总共四种情况。用二分实现即可，时间复杂度\(O(QLogC_l+QLogC_e)\)。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 100005; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int func(int dist, int v) { return dist / v + (dist % v != 0); } int a[MAXN], b[MAXN]; int main() { int r, n, m, v; read(r), read(r), read(n), read(m), read(v); for (int i = 1; i <= n; i++) read(a[i]); for (int i = 1; i <= m; i++) read(b[i]); int q; read(q); for (int i = 1; i <= q; i++) { int sx, sy, tx, ty; read(sx), read(sy); read(tx), read(ty); if (sx == tx) { writeln(abs(sy - ty)); continue; } int tmp = lower_bound(a + 1, a + n + 1, sy) - a; int tnp = tmp - 1; int ans = 2e9; if (tmp <= n) chkmin(ans, abs(a[tmp] - sy) + abs(a[tmp] - ty) + abs(sx - tx)); if (tnp >= 1) chkmin(ans, abs(a[tnp] - sy) + abs(a[tnp] - ty) + abs(sx - tx)); tmp = lower_bound(b + 1, b + m + 1, sy) - b; tnp = tmp - 1; if (tmp <= m) chkmin(ans, abs(b[tmp] - sy) + abs(b[tmp] - ty) + func(abs(sx - tx), v)); if (tnp >= 1) chkmin(ans, abs(b[tnp] - sy) + abs(b[tnp] - ty) + func(abs(sx - tx), v)); writeln(ans); } return 0; }【Div.2 D/Div.1 B】Resource Distribution
【思路要点】
由于一组服务器中服务器各自的功率为所有服务器的最小值，我们应当为两个需求分别选取在功率值域上连续的一段服务器。对功率排序，枚举被分配到功率较高的服务器的需求，默认其占用了最后的若干台服务器，再检查是否能够用剩余服务器满足另一个需求即可。时间复杂度\(O(NLogN)\)。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 300005; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct info {int val, home; }; info a[MAXN]; bool cmp(info a, info b) {return a.val < b.val; } int main() { int n, x, y; read(n), read(x), read(y); for (int i = 1; i <= n; i++) read(a[i].val), a[i].home = i; sort(a + 1, a + n + 1, cmp); int posx = 0, posy = 0; for (int i = n; i >= 1; i--) { int cnt = x / a[i].val + (x % a[i].val != 0); if (i + cnt - 1 <= n) { posx = i; break; } } if (posx == 0) { printf("No\n"); return 0; } for (int i = n; i >= 1; i--) { int cnt = y / a[i].val + (y % a[i].val != 0); if (i + cnt - 1 <= n) { posy = i; break; } } if (posy == 0) { printf("No\n"); return 0; } for (int i = 1; i <= n; i++) { int cntx = x / a[i].val + (x % a[i].val != 0); if (i + cntx <= posy) { printf("Yes\n%d %d\n", cntx, n - posy + 1); for (int j = 1; j <= cntx; j++) printf("%d ", a[i + j - 1].home); printf("\n"); for (int j = posy; j <= n; j++) printf("%d ", a[j].home); printf("\n"); return 0; } int cnty = y / a[i].val + (y % a[i].val != 0); if (i + cnty <= posx) { printf("Yes\n%d %d\n", n - posx + 1, cnty); for (int j = posx; j <= n; j++) printf("%d ", a[j].home); printf("\n"); for (int j = 1; j <= cnty; j++) printf("%d ", a[i + j - 1].home); printf("\n"); return 0; } } printf("No\n"); return 0; }【Div.2 E/Div.1 C】Big Secret
【思路要点】
任意排列小于\(2^i\)的数，不会对第\(i\)以及更高位的合法性产生影响。因此，在存在一个合法解的基础上，我们可以对其进行按位重排，使得其可以被下面的贪心算法找到：令\(s_0=0\)，若不存在使得\(s_{i-1}\ xor\ b_i>s_{i-1}\)成立的\(b_i\)，问题无解。否则在数集中找到使得\(s_{i-1}\ xor\ b_i>s_{i-1}\)成立的基础上\(s_{i-1}\ xor\ b_i\)最小的\(b_i\)，作为答案的下一个数，在数集中删除\(b_i\)，并令\(s_i=s_{i-1}\ xor\ b_i\)。暴力模拟上述贪心，我们可以得到\(O(N^2)\)的时间复杂度。用字典树加速上述贪心，时间复杂度降至\(O(NLog^2B_i)\)，下面笔者的代码实现的是这种做法。题解给出了一种复杂度更为优秀的做法。考虑我们已经完成了除了1以外所有数的排布，那么我们只需要将每个1放在任意的偶数后即可，若偶数数量不足，则无解。上文提到任意排列小于\(2^i\)的数，不会对第\(i\)以及更高位的合法性产生影响。注意到上述过程并不仅限于1，若我们已经完成了最高位大于\(i\)的数的排布，我们现在需要安排最高位为\(i\)的数，那么我们只主要将它们依次安排在第\(i\)位为0的位置后即可，若这样的位置不足，则无解。对所有数按最高位分类，用链表实现上述做法，时间复杂度\(O(NLogB_i)\)。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 100005; const int MAXLOG = 61; const int MAXP = 1e7 + 5; const long long INF = 9e18; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct SegmentTree { struct Node { int lc, rc; int sum; } a[MAXP]; int root, size; void insert(int &root, int depth, long long x, int d) { if (root == 0) root = ++size; a[root].sum += d; if (depth == -1) return; long long tmp = 1ll << depth; if (tmp & x) insert(a[root].rc, depth - 1, x, d); else insert(a[root].lc, depth - 1, x, d); } void insert(long long x, int d) { insert(root, MAXLOG, x, d); } long long query(int root, int depth, long long x, bool type) { if (depth == -1) return 0; long long tmp = 1ll << depth; if (type) { if (x & tmp) { if (a[a[root].rc].sum) return query(a[root].rc, depth - 1, x, type) + tmp; else return query(a[root].lc, depth - 1, x, type); } else { if (a[a[root].lc].sum) return query(a[root].lc, depth - 1, x, type); else return query(a[root].rc, depth - 1, x, type) + tmp; } } else { if (x & tmp) { if (a[a[root].lc].sum) return query(a[root].lc, depth - 1, x, type); else return -INF; } else { if (a[a[root].lc].sum) { long long tnp = query(a[root].lc, depth - 1, x, type); if (tnp > 0) return tnp; } if (a[a[root].rc].sum) return query(a[root].rc, depth - 1, x, true) + tmp; else return -INF; } } } long long query(long long x) { long long ans = query(root, MAXLOG, x, false); if (ans > 0) insert(ans, -1); return ans; } } ST; long long ans[MAXN]; int main() { int n; read(n); long long now = 0; for (int i = 1; i <= n; i++) { long long x; read(x); ST.insert(x, 1); } for (int i = 1; i <= n; i++) { long long tmp = ST.query(now); if (tmp <= 0) { printf("No\n"); return 0; } now ^= tmp; ans[i] = tmp; } printf("Yes\n"); for (int i = 1; i <= n; i++) write(ans[i]), putchar(' '); return 0; }【Div.2 F/Div.1 D】Aztec Catacombs
【思路要点】
对原图进行BFS，若起点到终点的最短路长度在3以内，输出该路径。引理1：在不存在长度为3的解的情况下，若存在解，一定存在一种最优解使得路径经过了一条原本不在图中的边。引理1证明：反证法：若最优解为一条起点到终点在原图上的长度大于等于4的最短路，那么最短路上的点除了相邻的点对以外一定两两在原图中没有边相连，否则可以构造出一条原图中更短的最短路。令最短路上前三个点为1、\(x\)、\(y\)，一定存在以下一条不劣于上述最优解的路径：1-\(x\)-\(y\)-1-\(N\)。枚举与1相邻的\(x\)，与\(x\)相邻的\(y\)，判断1和\(y\)间是否存在边，若不存在，则找到了一条长度为4的路径，将其输出。引理2：在不存在长度为4的解的情况下，若存在解，一定存在一种最优解形如：1-\(x\)-\(y\)-\(z\)-\(x\)-\(N\)，长度为5。引理2证明：首先，由引理1，最优解一定经过了一条原本不在图中的边。令这条边为\(z\)-\(x\)，也就是说在经过这条边之前，点\(x\)已经已经被经过了一次，若\(x\)原本与\(N\)相邻，那么最优解显然应当在第一次经过\(x\)时直接走向\(N\)，因此\(x\)原本与\(N\)不相邻，因此最优解的最后三个点必然为\(z\)-\(x\)-\(N\)，并且\(z\)-\(x\)是其经过的唯一一条原本不在图中的边。考虑最优解在经过\(z\)之前经过的每个点，它们都应当与1在原图中相邻，否则我们一定可以构造出一条长度为4的解。并且除去1以外，这些点之间应该除了相邻的点对以外一定两两在原图中没有边相连，否则可以构造出一条原图中更短的到达\(z\)点的路。因此，令最优解的前四个点为1、\(x\)、\(y\)、\(z\)，一定存在以下一条不劣于上述最优解的路径：1-\(x\)-\(y\)-\(z\)-\(x\)-\(N\)。由引理2证明，我们也发现，我们可以将于1相邻的，并且删去点1依然连通的一系列点一并考虑。对于每一系列这样的点，若它们形成了一个完全图，那么我们无法找到上述长度为5的路径。否则，找到一个度数不足点数的点作为\(x\)，枚举与\(x\)相邻的\(y\)、与\(y\)相邻的\(z\)，判断\(x\)和\(z\)间是否存在边，若不存在，则找到了一条长度为5的路径，将其输出。否则，问题无解。时间复杂度\(O(MLogN)\)。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 300005; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } vector <int> a[MAXN], pos; set <int> b[MAXN]; int n, m, degree[MAXN]; bool vis[MAXN]; void bfs() { static int fa[MAXN], q[MAXN], depth[MAXN]; static bool vis[MAXN]; int l, r; q[l = r = 1] = 1, depth[1] = 1, vis[1] = true; while (l <= r) { int tmp = q[l++]; for (unsigned i = 0; i < a[tmp].size(); i++) if (!vis[a[tmp][i]]) { q[++r] = a[tmp][i]; fa[a[tmp][i]] = tmp; vis[a[tmp][i]] = true; depth[a[tmp][i]] = depth[tmp] + 1; } } if (depth[n] != 0 && depth[n] <= 5) { int pos = n; vector <int> ans; ans.push_back(n); while (pos != 1) { pos = fa[pos]; ans.push_back(pos); } writeln(ans.size() - 1); for (unsigned i = ans.size(); i >= 1; i--) printf("%d ", ans[i - 1]); exit(0); } for (int i = 1; i <= r; i++) { int pos = q[i]; if (depth[pos] == 3 && b[pos].count(1) == 0) { writeln(4); printf("%d %d %d %d %d\n", 1, fa[pos], pos, 1, n); exit(0); } } } void work(int x) { vis[x] = true; pos.push_back(x); for (unsigned i = 0; i < a[x].size(); i++) { degree[a[x][i]]++; if (!vis[a[x][i]]) work(a[x][i]); } } int main() { read(n), read(m); for (int i = 1; i <= m; i++) { int x, y; read(x), read(y); a[x].push_back(y); a[y].push_back(x); b[x].insert(y); b[y].insert(x); } bfs(); vis[1] = true; for (unsigned i = 0; i < a[1].size(); i++) { if (vis[a[1][i]]) continue; pos.clear(); work(a[1][i]); int x = 0; for (unsigned j = 0; j < pos.size(); j++) if (degree[pos[j]] < pos.size() - 1) x = pos[j]; if (x == 0) continue; for (unsigned j = 0; j < a[x].size(); j++) { if (a[x][j] == 1) continue; int y = a[x][j]; for (unsigned k = 0; k < a[y].size(); k++) { int z = a[y][k]; if (z != x && z != 1 && b[x].count(z) == 0) { writeln(5); printf("%d %d %d %d %d %d\n", 1, x, y, z, x, n); exit(0); } } } } writeln(-1); return 0; }【Div.1 E】May Holidays
【思路要点】
直接用数据结构难以维护，考虑分块。将每\(K\)个操作分为一块，对询问涉及的点建立虚树，显然，每个操作只会影响原树上被虚树上的点夹住的整段。对每个整段进行排序+去重，即可在\(O(1)\)的时间内对其进行整体加1/减1，并动态维护答案。时间复杂度\(O(\frac{M}{K}(K^2+NLogN))\)，若排序使用基数排序，复杂度则为\(O(\frac{M}{K}(K^2+N))\)。笔者的做法直接使用排序，取\(K=\sqrt{NLogN}\)时，可得渐进意义下最优时间复杂度\(O(M\sqrt{NLogN})\)，常数很小。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 100005; const int MAXLOG = 20; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } vector <int> a[MAXN], b[MAXN], c[MAXN]; int n, m, father[MAXN], val[MAXN], q[MAXN]; int timer, dfn[MAXN], depth[MAXN], parent[MAXN][MAXLOG]; int btot, bsize, bl[MAXN], br[MAXN], belong[MAXN]; int ans, size[MAXN], newfa[MAXN], point[MAXN]; bool flg[MAXN]; int lca(int x, int y) { if (depth[x] < depth[y]) swap(x, y); for (int i = MAXLOG - 1; i >= 0; i--) if (depth[parent[x][i]] >= depth[y]) x = parent[x][i]; if (x == y) return x; for (int i = MAXLOG - 1; i >= 0; i--) if (parent[x][i] != parent[y][i]) { x = parent[x][i]; y = parent[y][i]; } return father[x]; } void dfs(int pos, int fa) { dfn[pos] = ++timer; parent[pos][0] = fa; for (int i = 1; i < MAXLOG; i++) parent[pos][i] = parent[parent[pos][i - 1]][i - 1]; depth[pos] = depth[fa] + 1; for (unsigned i = 0; i < a[pos].size(); i++) dfs(a[pos][i], pos); } void calc(int pos) { for (unsigned i = 0; i < a[pos].size(); i++) { calc(a[pos][i]); size[pos] += size[a[pos][i]]; } val[pos] += size[pos]; if (!flg[pos] && val[pos] <= 0) ans++; } bool cmp(int x, int y) { return dfn[x] < dfn[y]; } int main() { read(n), read(m); for (int i = 2; i <= n; i++) { read(father[i]); a[father[i]].push_back(i); } for (int i = 1; i <= n; i++) read(val[i]), val[i]++; for (int i = 1; i <= m; i++) read(q[i]); dfs(1, 0); bsize = pow(m, 0.618); btot = 0; for (int i = 1; i <= m; i++) { if (i % bsize == 1 % bsize) bl[++btot] = i; belong[i] = btot, br[btot] = i; } for (int t = 1; t <= btot; t++) { static int tmp[MAXN]; int tot = 0; for (int i = bl[t]; i <= br[t]; i++) tmp[++tot] = abs(q[i]); sort(tmp + 1, tmp + tot + 1, cmp); static int stk[MAXN], used[MAXN]; int top = 0, cnt = 0; stk[++top] = used[++cnt] = 1; for (int i = 1; i <= tot; i++) { if (tmp[i] == stk[top]) continue; int Lca = lca(tmp[i], stk[top]); if (Lca == stk[top]) stk[++top] = used[++cnt] = tmp[i]; else { while (dfn[Lca] < dfn[stk[top - 1]]) { newfa[stk[top]] = stk[top - 1]; top--; } newfa[stk[top]] = Lca; top--; if (Lca == stk[top]) stk[++top] = used[++cnt] = tmp[i]; else { stk[++top] = used[++cnt] = Lca; stk[++top] = used[++cnt] = tmp[i]; } } } while (top >= 2) { newfa[stk[top]] = stk[top - 1]; top--; } for (int i = 1; i <= cnt; i++) { int now = used[i], pos = father[now]; b[now].clear(), c[now].clear(); while (pos != newfa[now]) { if (!flg[pos]) b[now].push_back(val[pos]); pos = father[pos]; } sort(b[now].begin(), b[now].end()); int top = -1; for (unsigned i = 0; i < b[now].size(); i++) if (i == 0 || b[now][i] != b[now][i - 1]) b[now][++top] = b[now][i], c[now].push_back(1); else c[now][top]++; b[now].resize(top + 1); point[now] = 0; for (unsigned i = 0; i < b[now].size(); i++) if (b[now][i] <= 0) point[now]++; else break; } memset(size, 0, sizeof(size)); for (int i = bl[t]; i <= br[t]; i++) { int now = abs(q[i]); if (q[i] >= 0) { if (val[now] + size[now] <= 0) ans--; flg[now] = true; int pos = now; while (pos) { size[pos]--; if (!flg[pos] && val[pos] + size[pos] == 0) ans++; if (point[pos] < b[pos].size() && b[pos][point[pos]] + size[pos] <= 0) ans += c[pos][point[pos]++]; pos = newfa[pos]; } } else { if (val[now] + size[now] + 1 <= 0) ans++; flg[now] = false; int pos = now; while (pos) { if (now != pos && !flg[pos] && val[pos] + size[pos] == 0) ans--; size[pos]++; if (point[pos] > 0 && b[pos][point[pos] - 1] + size[pos] > 0) ans -= c[pos][--point[pos]]; pos = newfa[pos]; } } printf("%d ", ans); } ans = 0; memset(size, 0, sizeof(size)); for (int i = bl[t]; i <= br[t]; i++) if (q[i] >= 0) size[q[i]]--, flg[q[i]] = true; else size[-q[i]]++, flg[-q[i]] = false; calc(1); } return 0; }【Div.1 F】Parametric Circulation
【思路要点】
按照有上下界的网络流建图，问题转化为\(S\)至\(T\)满流（流量为\(\sum_{e\in E}l_e\)）的概率。令\(f(x)\)为当\(t=x\)时\(S\)至\(T\)的最大流，我们接下来要证明这个函数是在\(x\in [0,1]\)上的上凸函数。以\(t\)的值为横坐标，\(S\)至\(T\)的流量为纵坐标，建立平面直角坐标系，由于各边的流量为关于\(t\)的线性函数，而网络流本质上是线性规划问题，因此该问题同样可以转化为线性规划问题，而线性规划问题的解空间具有凸性，因此若\((x,y)\)和\((a,b)\)是可以达到的状态（即当\(t=x\)，存在\(S\)至\(T\)的大小为\(y\)的可行流；当\(t=a\)，存在\(S\)至\(T\)的大小为\(b\)的可行流），那么\((x,y)\)与\((a,b)\)的连线上一定也都是可以达到的状态，由此，我们可以直观地证明这个函数的上凸性。而显然\(\sum_{e\in E}l_e\)为一个一次函数。因此，函数\(g(x)=\sum_{e\in E}l_e-f(x)\)在\(x\in [0,1]\)上是下凸函数，并且由\(g(x)\)的定义显然有\(g(x)≥0\)，我们关心的则是\(g(x)\)取值为0的范围。三分求出\(g(x)\)的最小值及其位置，再向两边二分求得其范围即可。时间复杂度\(O(Dinic(N,M)*LogV)\)，其中\(V\)为精度要求。注意由于精度要求有限，我们可以将所有数乘以\(10^8\)，将问题转化为整数运算的问题。【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2005; const long long bnd = 1e8; const long long INF = 1e18; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct edge {int dest; long long flow; unsigned home; }; vector <edge> a[MAXN]; int s, t, n, m, x[MAXN], y[MAXN], dist[MAXN]; long long lk[MAXN], lb[MAXN], rk[MAXN], rb[MAXN]; unsigned curr[MAXN]; long long dinic(int pos, long long limit) { if (pos == t) return limit; long long used = 0; for (unsigned &i = curr[pos]; i < a[pos].size(); i++) if (a[pos][i].flow != 0 && dist[pos] + 1 == dist[a[pos][i].dest]) { long long tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow)); used += tmp; a[pos][i].flow -= tmp; a[a[pos][i].dest][a[pos][i].home].flow += tmp; if (limit == used) return used; } return used; } bool bfs() { static int q[MAXN], l, r; for (int i = 0; i <= t; i++) dist[i] = 0; dist[s] = 1, q[l = r = 0] = s; while (l <= r) { int tmp = q[l++]; for (unsigned i = 0; i < a[tmp].size(); i++) if (dist[a[tmp][i].dest] == 0 && a[tmp][i].flow != 0) { dist[a[tmp][i].dest] = dist[tmp] + 1; q[++r] = a[tmp][i].dest; } } return dist[t] != 0; } void addedge(int s, int t, long long x) { a[s].push_back((edge) {t, x, a[t].size()}); a[t].push_back((edge) {s, 0, a[s].size() - 1}); } void addedge(int x, int y, long long l, long long r) { addedge(s, y, l); addedge(x, y, r - l); addedge(x, t, l); } long long f(long long val) { long long diff = 0; s = 0, t = n + 1; for (int i = s; i <= t; i++) a[i].clear(); for (int i = 1; i <= m; i++) { addedge(x[i], y[i], lk[i] * val + lb[i], rk[i] * val + rb[i]); diff += lk[i] * val + lb[i]; } while (bfs()) { memset(curr, 0, sizeof(curr)); diff -= dinic(s, INF); } return diff; } int main() { read(n), read(m); for (int i = 1; i <= m; i++) { read(x[i]), read(y[i]); read(lk[i]), read(lb[i]), lb[i] *= bnd; read(rk[i]), read(rb[i]), rb[i] *= bnd; } long long l = 0, r = bnd; while (l + 5 < r) { long long ml = (2 * l + r) / 3; long long mr = (l + 2 * r) / 3; if (f(ml) < f(mr)) r = mr; else l = ml; } long long mid = l; for (int i = l; i <= r; i++) if (f(i) == 0) mid = i; l = 0, r = mid; while (l < r) { long long md = (l + r) / 2; if (f(md) == 0) r = md; else l = md + 1; } long long ql = l; l = mid, r = 1; while (l < r) { long long md = (l + r + 1) / 2; if (f(md) == 0) l = md; else r = md - 1; } long long qr = l; printf("%.10lf\n", (double) (qr - ql) / bnd); return 0; }