HDU-2616

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1525    Accepted Submission(s): 1043

Problem Description

There is a mountain near yifenfei’s hometown. On the mountain lived a big monster. As a hero in hometown, yifenfei wants to kill it. 
Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect.

 

 

Input

The input contains multiple test cases.
Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has.
Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m).

 

 

Output

For each test case output one integer that how many spells yifenfei should use at least. If yifenfei can not kill the monster output -1.

 

 

Sample Input

3 100

10 20

45 89

5 40

 

3 100

10 20

45 90

5 40

 

 

3 100

10 20

45 84

5 40

 

 

Sample Output

3

2

-1

 

题意：

有n个符咒，怪兽有m点血。每个符咒有两个属性：

1.伤害A；

2.在怪物血量低于M时造成2*A点伤害。

 

AC代码：

#include<bits/stdc++.h>
using namespace std;

struct node{
    int x,y;
}a[11];

bool vis[11];
int n,m,ans;

void dfs(int m,int len){
    if(len>=ans)
        return;
    if(m<=0){
        ans=min(ans,len);
        return;
    }
    for(int i=0;i<n;i++){
        if(vis[i]==0){
            vis[i]=1;
            if(m<=a[i].y)
            dfs(m-2*a[i].x,len+1);
            else
            dfs(m-a[i].x,len+1);
            vis[i]=0;
        }
    }
}

int main(){
    while(cin>>n>>m){
        for(int i=0;i<n;i++){
            cin>>a[i].x>>a[i].y;
        } 
        memset(vis,0,sizeof(vis));
        ans=11;
        dfs(m,0);
        if(ans>10){
            cout<<-1<<endl;
        }
        else{
            cout<<ans<<endl;
        }
    }
    return 0;
}