A - Kill the monster （DFS）

水题

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

#include<stdio.h>
#include<algorithm>
using namespace std;
#include<iostream>
#include<string.h>
int A,M,MIN;
int vis[20],AI[20],MI[20];
void DFS(int blood,int num)
{
    if(num>A)//技能次数超过技能数
    {
        return ;
    }
    if(blood<=0)
    {
        if(num<MIN)
            MIN=num;
        return ;
    }
    for(int i=0; i<A; i++)
    {
        if(!vis[i])
        {
            vis[i]=1;//标记用过的技能
            if(blood<=MI[i])
                DFS(blood-2*AI[i],num+1);//暴击效果
            else
                DFS(blood-AI[i],num+1);//普攻
            vis[i]=0;
        }
    }
}
int main()
{
    while(~scanf("%d%d",&A,&M))
    {
        for(int i=0; i<A; i++)
        {
            scanf("%d%d",&AI[i],&MI[i]);
            vis[i]=0;
        }
        MIN=A+1;
        DFS(M,0);//血量和使用次数
        if(MIN==A+1)
            printf("-1\n");
        else
            printf("%d\n",MIN);
    }
}