题目大意：
输入n, w, m 代表有n个桶，每桶w升牛奶，m个杯子等待分配牛奶，要求没个杯子中的牛奶量相同，且每个桶至多给2个杯子分配牛奶，若可以分配，输出YES，和个杯子是由第几个桶 分配了多少升，若不能分配输出NO。
思路：
用贪心的思想将每个桶中的牛奶每次最大量的分出，用vector<结构体>记录下分配过程，此外由于可能出现除不尽的情况，因此还要控制下精度，使用fabs（a-b）与1e-6比较即可。
ac代码：
#include <iostream> #include <vector> #include <cmath> #define pause system("pause") #define cout(x) cout << fixed << setprecision(x) using namespace std; const int maxn = 305; int n ,w, m; typedef struct node { double w; int n; node(double ww, int nn) { w = ww; n = nn; } } node; double wi[maxn]; int use[maxn]; double mi[maxn]; vector<node>Get[maxn]; double avg; void init() { for (int i = 1; i <= max(m, n); i++) { use[i] = 0; wi[i] = (double) w; mi[i] = 0; Get[i].clear(); } } bool Fill(int inx) { //cout << "inx = " << inx << endl; for (int i = 1; i <= n; i++) { if (wi[i] == 0) continue; if (fabs(wi[i] + mi[inx] - avg) <= 1e-6) { //cout << 1 << endl; Get[inx].push_back(node(wi[i], i)); wi[i] = 0; mi[inx] = avg; } else if (wi[i] > avg - mi[inx]) { //cout << 2 << endl; wi[i] -= (avg - mi[inx]); Get[inx].push_back(node(avg - mi[inx], i)); mi[inx] = avg; } else { //cout << 3 << endl; Get[inx].push_back(node(wi[i], i)); mi[inx] += wi[i]; wi[i] = 0; } use[i]++; if (use[i] > 2) return false; if (fabs(avg - mi[inx]) <= 1e-6) return true; } return true; } int main() { cin >> n >> w >> m; if (m > 2 * n) cout << "NO" << endl; else { if (n == m) { cout << "YES" << endl; for (int i = 1; i <= n; i++) cout << i << " " << w << endl; } else { avg = (double) (n * w) / m; init(); int flag = 1; //if (n < m) { for (int i = 1; i <= m; i++) { bool f = Fill(i); if (!f) { flag = 0; break; } } if (!flag) cout << "NO" << endl; else { cout << "YES" << endl; for (int u = 1; u <= m; u++) { for (int i = 0; i < Get[u].size(); i++) { if (i) cout << " "; printf("%d %.6lf", Get[u][i].n, Get[u][i].w); } cout << endl; } } //} } } return 0; }
