hdu2616 kill the monster 4.3.8

据说是道dfs水题……

Kill the monster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 41 Accepted Submission(s): 34 

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 10010 2045 895  403 10010 2045 905 403 10010 2045 845 40

 

Sample Output

32-1

 

#include <iostream>using namespace std;int n,m;int a[12],b[12];int hash[12];int maxx;void dfs(int s,int m)//number of used spell, remained hp{     int i;     if(s>n)//the number of spell is over         return;     if(m<=0)//hp<0     {         if(maxx>s)//find a smaller method             maxx=s;         return;     }          for(i=0;i<n;i++)     {         if(hash[i])//didn't use this spell         {             hash[i]=0;//now use it             if(m<=b[i])//if left hp is less than double-effect hp             {                 dfs(s+1,m-2*a[i]);             }             else                 dfs(s+1,m-a[i]);             hash[i]=1;//recovery         }     }     return ;}                                       int main(){    int i;    while(cin>>n>>m)    {        for(i=0;i<n;i++)        {            cin>>a[i]>>b[i];        }        memset(hash,1,sizeof(hash));        maxx=15;        dfs(0,m);        if(maxx==15)        cout<<"-1"<<endl;        else        cout<<maxx<<endl;//Presentation Error....shame    }    return 0;}

大家都是怎么看出来什么时候用dfs的呢……