题目链接：
题意：给出每个任务完成需要的时间t和要求的截止时间d。求最多可以完成多少个？
思路：按照截止时间升序排序。对于每个任务：（1）若之前耗费总时间加上t小于等于d，那么当前任务可以完成，插入优先队列（以t的大小作为关键字）；（2）若不能，比较队头元素的时间和当前的时间t。若大于当前时间，则踢掉队头元素插入当前任务。因为我们是按照d升序的，所以对于以后的任务，若之前完成需要的总时间越少，那么后面的任务越有可能完成。 最后队里元素个数就是答案。
#include <iostream>
#include <cstdio>#include <string.h>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <set>#include <stack>#include <string>#include <map>#define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))#define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define FOR0(i,x) for(i=0;i<x;i++)#define FOR1(i,x) for(i=1;i<=x;i++)#define FOR(i,a,b) for(i=a;i<=b;i++)#define FORL0(i,a) for(i=a;i>=0;i--)#define FORL1(i,a) for(i=a;i>=1;i--)#define FORL(i,a,b)for(i=a;i>=b;i--)#define rush() int CC;for(scanf("%d",&CC);CC--;)#define Rush(n) while(scanf("%d",&n)!=-1)using namespace std;void RD(int &x){scanf("%d",&x);}void RD(i64 &x){scanf("%I64d",&x);}void RD(u64 &x){scanf("%I64u",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(i64 &x,i64 &y){scanf("%I64d%I64d",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(i64 &x,i64 &y,i64 &z){scanf("%I64d%I64d%I64d",&x,&y,&z);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;}void PR(int x) {printf("%d\n",x);}void PR(int x,int y) {printf("%d %d\n",x,y);}void PR(i64 x) {printf("%I64d\n",x);}void PR(u32 x) {printf("%u\n",x);}void PR(u64 x) {printf("%I64u\n",x);}void PR(double x) {printf("%.6lf\n",x);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout<<x<<endl;}const int mod=100003;const i64 inf=((i64)1)<<60;const double dinf=1e50;const int INF=1000000000;const int N=150005;struct node{ int t,d; void get() { RD(t,d); } int operator<(const node &a) const { return t<a.t; }};int n;node a[N];priority_queue<node> Q;int cmp(node a,node b){ return a.d<b.d;}int main(){ RD(n); int i; FOR1(i,n) a[i].get(); sort(a+1,a+n+1,cmp); int cur=0; node temp; FOR1(i,n) { if(cur+a[i].t<=a[i].d) { Q.push(a[i]); cur+=a[i].t; } else if(!Q.empty()) { temp=Q.top(); if(temp.t>a[i].t) { cur=cur-temp.t+a[i].t; Q.pop(); Q.push(a[i]); } } } PR(SZ(Q)); return 0;}