题意:有n(n<=800000)个工作,已知每个工作需要的时间qi和截止时间di(必须在此之前完成),最多能完成多少个工作?工作只能串行完成。第一项任务开始的时间不早于时刻0。
分析:按截止时间排序,在若当前工作完成后超过截止时间,则优先选择需要时间少的工作,即若优先队列中需要时间最长的工作a,
1、其时间大于当前工作b,则选择完成工作b,弹出a。(此操作的前提是完成b弹出a后的时间要小于等于b的截止时间)
2、其时间小于当前工作b,则没必要弹出a。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) const double eps = 1e-8; inline int dcmp(double a, double b) { if(fabs(a - b) < eps) return 0; return a < b ? -1 : 1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 800000 + 10; const int MAXT = 10000 + 10; using namespace std; struct Node{ int q, d; void read(){ scanf("%d%d", &q, &d); } bool operator < (const Node& rhs)const{ return d < rhs.d; } }num[MAXN]; int n; int solve(){ priority_queue<int> q; int st = 0; for(int i = 0; i < n; ++i){ if(st + num[i].q <= num[i].d){ st += num[i].q; q.push(num[i].q); } else if(!q.empty()){ int tmp = q.top(); if(tmp > num[i].q && st - tmp + num[i].q <= num[i].d){ q.pop(); q.push(num[i].q); st += num[i].q - tmp; } } } return q.size(); } int main(){ int T; scanf("%d", &T); while(T--){ scanf("%d", &n); for(int i = 0; i < n; ++i) num[i].read(); sort(num, num + n); printf("%d\n", solve()); if(T) printf("\n"); } return 0; }