HDU 2616 Kill the monster(简单DFS)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2616

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 594    Accepted Submission(s): 417

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

3 100 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

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

 

 

简单DFS搜索题。。还好1A了，不然真弱爆了。。。。

 

 By LFENG

#include<stdio.h> #include<string.h> struct spell {     int effect;     int hp; }sp[30]; int n,m,mint,HP; int vis[30]; void getdata() {     int i;     for(i=0;i<n;i++)     scanf("%d %d",&sp[i].effect,&sp[i].hp);     mint=9999999;     memset(vis,0,sizeof(vis)); } void dfs(int num,int mhp) {     int i;     if(mhp<=0)     {         if(mint>num)mint=num;         return ;     }     for(i=0;i<n;i++)     {         if(vis[i])continue;         vis[i]=1;         if(mhp<=sp[i].hp)dfs(num+1,mhp-sp[i].effect*2);         else dfs(num+1,mhp-sp[i].effect);         vis[i]=0;     } } int main() {     int t;     while(scanf("%d %d",&n,&HP)!=EOF)     {         getdata();         dfs(0,HP);         if(mint<9999999)printf("%d\n",mint);         else printf("-1\n");     }     return 0; }