题目：
Kill the monster
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1404 Accepted Submission(s): 959
**
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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.\
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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).\
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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.
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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
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Sample Output
3
2
-1
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Author
yifenfei
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Source
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Recommend
yifenfei
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Statistic | Submit | Discuss | Note
思路：
一个人要打BOSS,他有n个技能，每个技能可以造成不同伤害，针对每一个技能，当BOSS的血量低于这个值时可以触发暴击导致2被伤害，题目问最少需要放几个技能可以打死BOSS，如果技能放完了打不死就输出-1，简单深搜。
代码：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int aa[20];//每个技能的伤害量
int mm[20];//低于这这个值可以双倍暴击
int vis[20];//标记
int n,m,step;
void dfs(int num,int s)
{
if(s>=step)return;//剪枝
if(num<=0)
{
if(s<step)step=s;//更新步数
return;
}
for(int i=0; i<n; i++)
{
if(!vis[i])
{
vis[i]=1;
if(num<=mm[i])//判断是否能暴击
dfs(num-2*aa[i],s+1);
else
dfs(num-aa[i],s+1);
vis[i]=0;
}
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
step=99;
mem(vis,0);
mem(aa,0);
mem(mm,0);
for(int i=0; i<n; i++)
scanf("%d%d",&aa[i],&mm[i]);
dfs(m,0);
if(step==99)
printf("-1\n");
else
printf("%d\n",step);
}
return 0;
}
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