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
#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;
}
hdu 2616 暴力使用 dfs求最短路径（剪枝有点依稀）
原文：http://www.cnblogs.com/z1141000271/p/5472294.html
