- Kill the monster HDU2616

首先要了解深搜的模板。dfs(所有情况),比如这题的两种情况，魔王的HP值是该减去一般情况所受伤害，还是该减去double伤害值呢？？？？在搜索过程，记录搜索次数即可。

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 
     
     
     
     

<pre name="code" class="cpp">#include<stdio.h>
int a[20],b[20],n,w,c[20],min;
void dfs(int w,int z)
{
    int i;
    if(z>n)
        return ;//攻击次数不能大于勇士所会的技能个数
    if(w<=0)
    {
        if(z<min)
            min=z;
        return ;
    }
    for(i=0; i<n; i++)
    {
        if(!c[i])
        {
            c[i]=1;//标记
            if(w<=b[i])
                dfs(w-2*a[i],z+1);//搜索
            else
                dfs(w-a[i],z+1);
            c[i]=0;
            <span style="font-family: Arial, Helvetica, sans-serif;">//用过的技能不能再用</span>

        }
    }
}
int main()
{
    int i;
    while(scanf("%d%d",&n,&w)!=EOF)
    {

        for(i=0; i<n; i++)
        {
            scanf("%d%d",&a[i],&b[i]);
            c[i]=0;
        }
        min=20;
        dfs(w,0);
        if(min==20)
            printf("-1\n");
        else
            printf("%d\n",min);

    }
    return 0;
}