zoj 3410 Layton's Escape

贪心：

 

1.先固定攻克trap的次序：按di升序排列

 

2.确定哪一步进行hp--：

   设置队列存放状态，按排好序的依次入队,队列中按ti从大到小排列，当遇到total time>di的时候就将队头元素（ti花费最大的出队），并给hp++，这样必然时刻保持hp花费最小的，所以hp达到k时跳出

 

#include<iostream>#include<algorithm>#include<queue>#include<stdio.h>using namespace std;#define N 25010struct Status{int ti;int di;bool operator <(const Status a)const{return ti<a.ti;}}trap[N],now,temp;int comp(Status a,Status b){return a.di<b.di;}int n,k;int bfs(){int i,j;int total=0;int hp=0;priority_queue<Status>Q;for(i=0;i<n;i++){now.di=trap[i].di;now.ti=trap[i].ti;if(total+now.ti<=now.di){total+=now.ti;Q.push(now);}else{total+=now.ti;Q.push(now);if(total>now.di){temp=Q.top();Q.pop();total-=temp.ti;hp++;if(hp==k)return -1;}}}return hp;}int main(){while(scanf("%d%d",&n,&k)!=EOF){for(int i=0;i<n;i++)scanf("%d%d",&trap[i].ti,&trap[i].di);sort(trap,trap+n,comp);printf("%d/n",bfs());}}