zoj 3410 Layton's Escape（贪心+优先队列）

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Layton's Escape

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Professor Layton is a renowned archaeologist from London's Gressenheller University. He and his apprentice Luke has solved various mysteries in different places.

Unfortunately, Layton and Luke are trapped in a pyramid now. To escape from this dangerous place, they need to pass N traps. For each trap, they can use Ti minutes to remove it. If they pass an unremoved trap, they will lose 1 HP. They have K HP at the beginning of the escape and they will die at 0 HP.

Of course, they don't want trigger any traps, but there is a monster chasing them. If they haven't pass the ith trap in Di minutes, the monster will catch and eat them. The time they start to escape is 0, and the time cost on running will be ignored. Please help Layton to escape from the pyramid with the minimal HP cost.

Input

There are multiple test cases (no more than 20).

For each test case, the first line contains two integers N and K (1 <= N <= 25000, 1 <= K <= 5000), then followed by N lines, the ith line contains two integers Ti and Di (0 <= Ti <= 10^9, 0 <= Di <= 10^9).

Output

For each test case, if they can escape from the pyramid, output the minimal HP cost, otherwise output -1.

Sample Input

3 240 6060 9080 1202 130 12060 40

Sample Output

1-1

题意：有一个人要通过金字塔，但是有很多陷阱，他必须过陷阱，每经过一个陷阱需要ti时间，但是后面还有一个怪物，在追着他们，如果总花费时间超过di，那么他们需要花费一点精力，当精力到0的时候，他们就死了，求这个人能不能通过金字塔。能通过是，最少需要多少精力。

贪心+优先队列，如果超过了时间，那么减掉最少的一定是最好的，这是需要用到优先队列。

#include<iostream>#include<cmath>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;struct point{    long long t;    long long d;}p[25002];bool cmp(point a,point b){   if(a.d==b.d)    return a.t<b.t;   else     return a.d<b.d;}int main(){   int i,j,n,k;   while(~scanf("%d%d",&n,&k))   {       priority_queue<int>q;       for(i=0;i<n;i++)       {           scanf("%lld%lld",&p[i].t,&p[i].d);       }       sort(p,p+n,cmp);       long long T=0;       int tt=0;    for(i=0;i<n;i++)     {        T+=p[i].t;        q.push(p[i].t);        //cout<<T<<"&&&"<<endl;         while(T>p[i].d)           {               tt++; T-=q.top();              q.pop();           }         //cout<<tt<<"^^"<<endl;     }     if(tt>=k)         printf("-1\n");     else         printf("%d\n",tt);   }    return 0;}