ZOJ 3410 Layton's Escape

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 2

40 60

60 90

80 120

2 1

30 120

60 40

Sample Output

1
-1

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Author: JIANG, Kai
Contest: ZOJ Monthly, October 2010 

按d排序，如果有陷阱过不去，就去掉用时最长的那个陷阱。。。。。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

using namespace std;

struct node

{

    int t,d;

    bool operator<(const node a) const

    {

        return t<a.t;

    }

}p[26000];

bool cmp(node a,node  b)

{

    return a.d<b.d;

}

int main()

{

    int n,k;

while(cin>>n>>k)

{

    for(int i=0;i<n;i++)

    {

        int x,y;

        cin>>x>>y;

        p.t=x;   p.d=y;

    }

    sort(p,p+n,cmp);

    priority_queue<node> pq;

    int ddt=0;

    int ans=0;

    for(int i=0;i<n;i++)

    {

        pq.push(p);

        ddt+=p.t;

        if(ddt>p.d)

        {

            node x=pq.top();

            ddt-=x.t;

            pq.pop();

            ans++;

        }

    }

    ans=ans>=k? -1:ans;

    cout<<ans<<endl;

}

    return 0;

}