hdu 2616 暴力使用 dfs求最短路径（剪枝有点依稀）

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1241    Accepted Submission(s): 846

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

 #include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int mapp[11][2];
int vis[11];
int minn,n,m,flag;
void dfs(int bloor,int time)
{
 int i,j;
 if(bloor<=0)
 {
 // cout<<".."<<endl;
  if(minn>time) minn=time;
 // cout<<maxx<<endl;
  flag=1;
  return;
 }
 if(time>=minn) return ;//  当次数比以前走过的次数多的时候 剪去
 for(i=1;i<=n;i++)
 {
  if(vis[i]==1) continue;
  vis[i]=1;
 // cout<<".."<<endl;
  if(bloor<=mapp[i][1])
  {
  // cout<<".."<<endl;
      dfs(bloor-2*mapp[i][0],time+1); 
     }
  else
  {
  // cout<<".."<<endl;
   dfs(bloor-mapp[i][0],time+1);
  }
  vis[i]=0;
 }
}
int main()
{
 int i,j;
 while(cin>>n>>m)
 {
  for(i=1;i<=n;i++)
  {
   scanf("%d %d",&mapp[i][0],&mapp[i][1]);
  // cout<<mapp[i][0]<<mapp[i][1]<<endl;
  }
  memset(vis,0,sizeof(vis));
  flag=0;
  minn=999;
  dfs(m,0);
  if(flag==0) printf("-1\n");
  else printf("%d\n",minn);
 }
 return 0;
}