hdu2616

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1546    Accepted Submission(s): 1060

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
题意分析：有N个技能去打HP有M的怪兽，技能（A，M），技能伤害为A，当怪兽HP<=M时伤害为2*A。求打死怪兽（HP<=0）用的最少技能
AC代码：
#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[15],b[15];int used[15];int n,m;int minf;void DFS(int blood,int cnt){        if(cnt>n)                return;        if(cnt<minf&&blood<=0)        {                minf=cnt;                return;        }        for(int i=1;i<=n;i++)        {                if(!used[i])                {                        used[i]=1;                        if(blood<=b[i])                                DFS(blood-2*a[i],cnt+1);                        if(blood>b[i])                                DFS(blood-a[i],cnt+1);                        used[i]=0;                }        }}int main(){        while(scanf("%d%d",&n,&m)!=EOF)        {                for(int i=1;i<=n;i++)                        scanf("%d%d",&a[i],&b[i]);                memset(used,0,sizeof(used));                minf=20;                DFS(m,0);                if(minf!=20)                        printf("%d\n",minf);                else                        printf("-1\n");        }        return 0;}