题目地址 题意：给你n个学生的编号（string），m门课的班级编号，班级容量以及上课时间，q个选课请求，问有多少个请求可以达成（两门课的时间不能重叠才能选）。选课顺序是要按照课程在input出现的顺序排序，相同课程的按请求出现的顺序排序。 思路：就直接模拟，要注意细节，我就是排序规则函数里面把id写成了sx，WA了一个小时，这种题目仔细就好了，详细看代码
#include <iostream> #include <cstring> #include <string> #include <queue> #include <vector> #include <map> #include <set> #include <stack> #include <cmath> #include <cstdio> #include <algorithm> #include <iomanip> #define N 200010 #define M 400010 #define LL __int64 #define inf 0x3f3f3f3f #define lson l,mid,ans<<1 #define rson mid+1,r,ans<<1|1 #define getMid (l+r)>>1 #define movel ans<<1 #define mover ans<<1|1 using namespace std; const LL mod = 1000000007; struct node { set<int> kc; set<int> time; }student; struct nope { int cap; set<int> time; }kc; struct nodp { string name; int id; int sx; }px[510]; map<int, nope> kcj;//课程集 map<string, node> xsj;//学生集 map<int, int >kcx;//课程序 bool cmp(nodp a, nodp b) { if (kcx[a.id] == kcx[b.id]) { return a.sx < b.sx; } return kcx[a.id] < kcx[b.id]; } int main() { cin.sync_with_stdio(false); int Case = 1; int n, m, q; int a, b, c; int ans; string str; while (cin >> n >> m >> q) { kcj.clear(); xsj.clear(); kcx.clear(); student.kc.clear(); student.time.clear(); for (int i = 1; i <= n; i++) { cin >> str; xsj[str] = student; } for (int i = 1; i <= m; i++) { kc.time.clear(); cin >> a; cin >> kc.cap; cin >> b; kcx[a] = i; for (int j = 0; j < b; j++) { cin >> c; kc.time.insert(c); } kcj[a] = kc; } for (int i = 0; i < q; i++) { cin >> px[i].name >> px[i].id; px[i].sx = i; } sort(px, px + q, cmp); ans = 0; for (int i = 0; i < q; i++) { str = px[i].name; a = px[i].id; kc = kcj[a]; student = xsj[str]; if (kc.cap <= 0 || student.kc.count(a) != 0) { continue; } b = 1; for (set<int> ::iterator it = kc.time.begin(); it != kc.time.end(); it++) { if (student.time.count(*it) != 0) { b = 0; break; } } if (b == 1) { ans++; kcj[a].cap--; xsj[str].kc.insert(a); for (set<int> ::iterator it = kc.time.begin(); it != kc.time.end(); it++) { xsj[str].time.insert(*it); } } } cout << "Case " << Case++ << ": " << ans << endl; } return 0; }