意甲冠军:略。
思考:馋。给了计算公式,得到的钱很特别的标题500*x+2*y,y<=100,根据任务时间的降序是可能,
然后每个任务,找出多任务和任务的难度难度已完成最近的机。
#include <set> #include <cstdio> #include <iostream> #include <algorithm> using namespace std; typedef long long LL; const int N = 100005; struct Task { int x, y; bool operator<(const Task &rhs) const { if(x == rhs.x) return y > rhs.y; return x > rhs.x; } } task[N]; int main() { int n, m; while(cin >> n >> m) { multiset<int> S[105]; LL ans = 0; int maxLevel = 0, count = 0; for(int i = 0; i < n; i++) { int x, y; scanf("%d%d", &x, &y); S[y].insert(x); maxLevel = max(maxLevel, y); } for(int i = 0; i < m; i++) { scanf("%d%d", &task[i].x, &task[i].y); } sort(task, task + m); for(int i = 0; i < m; i++) { int x = task[i].x; int y = task[i].y; for(int j = y; j <= maxLevel; j++) { if(S[j].empty()) continue; multiset<int>::iterator iter = S[j].lower_bound(x); if(iter == S[j].end() || *iter < x) { continue; } else { count++; ans += 500ll * x + 2ll * y; S[j].erase(iter); break; } } } cout << count << " " << ans << endl; } return 0; }
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时间: 2024-10-10 16:34:55