HDU2616 Kill the monster(深搜DFS)

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

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

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

 

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;}