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.

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).

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.

3 100
10 20
45 89
5  40

3 100
10 20
45 90
5 40

3 100
10 20
45 84
5 40

3
2
-1


题意：给定技能次数和怪物血量，在n次下打死怪物，规定：技能的AI为攻击伤害,MI为如果怪物的血量小于MI，此处的攻击伤害加倍。


解题思路：深搜，cot 用于递归技能次数，m为怪物的血量，每次递归cot+1,m减去相应的血量，当m<=0时即为打死怪物，此时比较寻找最少技能次数。最终返回一个最小次数。


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int n,m,a[11][2],cot,visit[11],sum,min;
int dfs(int cot,int m)//cot记录攻击次数，m为怪物血量
{
    int i;
    if(m <= 0)//当怪物血量为零时，寻找最小次数
    {
        min = min > cot ? cot : min;
    }
    for(i=1;i<=n;i++)//遍历数组
    {
        if(!visit[i])//是否用过
        {
             visit[i] = 1;//标记
            if(m <= a[i][1])//如果怪物血量小于技能的MI值，攻击伤害为2倍
                dfs(cot+1,m-2*a[i][0]);//所以递归将血量减去2倍的攻击伤害
            else
               dfs(cot+1,m-a[i][0]);//否则减去相应的血量即可
           visit[i]  = 0;//回溯
        }
    }
return min;}
int main()
{
    int i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(visit,0,sizeof(visit));
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i][0],&a[i][1]);
        }
        min = 9999;
        if(dfs(0,m) == 9999)//如果min没有改变说明所有的技能都用完怪物的血量没有为0，即输出-1
            printf("-1\n");
        else
            printf("%d\n",min);//否则输出深搜得出的最小次数即可
    }
    return 0;
}