杭电acm 2616
时间：2014-04-14 02:22:38
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Kill the monster
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 818 Accepted Submission(s): 588
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.
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).
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>
int n,a[11],b[11],c[11],min;
void dfs(int cur,int hp)
{
int i;
if(hp<=0)
if(cur<min)
min=cur;
if(cur>n)
;
for(i=0;i<n;i++)
{
if(!c[i])
{
if(hp>b[i])
{
c[i]=1;
dfs(cur+1,hp-a[i]);
c[i]=0;
}
else
{
c[i]=1;
dfs(cur+1,hp-2*a[i]);
c[i]=0;
}
}
}
}
int main()
{
int i,m;
while(~scanf("%d%d",&n,&m))
{
memset(c,0,sizeof(c));
for(i=0;i<n;i++)
scanf("%d%d",&a[i],&b[i]);
min=11;
dfs(0,m);
if(min!=11)
printf("%d\n",min);
else
printf("-1\n");
}
return 0;
}
原文：http://blog.csdn.net/fanerxiaoqinnian/article/details/23620567
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