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原创于 2017-06-24 13:31:07 发布 · 310 阅读

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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

Sample Output

3
2
-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;
}