﻿HDU 2616 Kill the monster(DFS)-Ocrosoft

# Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1289    Accepted Submission(s): 880

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

```

Author
yifenfei

Source

```#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <conio.h>
#include <cmath>
#include <algorithm>
#include <list>
#include <stack>
#include <string>
#include <vector>
#include <cstring>
#include <cctype>
#include <queue>
#include <set>
#include <climits>
#define ms(a) memset(a,0,sizeof(a))
using namespace std;
template <typename T>
{
cin>>t;
return t;
}
int n,m;//skill,health
int ans;
struct S
{
int a;//attack
int hd;//health-double
} s[12];
bool use[12];
void dfs(int h,int k)
{
if(k>=ans)return;
if(h<=0)
{
ans=min(ans,k);
return;
}
for(int i=0; i<n; i++)
{
if(use[i])continue;
use[i]=1;
dfs(h-(h<=s[i].hd?2*s[i].a:s[i].a),k+1);
use[i]=0;
}
}
int main()
{
while(cin>>n>>m)
{
for(int i=0; i<n; i++)cin>>s[i].a>>s[i].hd;
ans=12;
ms(use);
dfs(m,0);
printf("%d\n",ans==12?-1:ans);
}
return 0;
}
```