hdu 2616 Kill the monster (DFS）

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 778    Accepted Submission(s): 556

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

 

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//109MS    228K    694 B    G++    姜伯约
/*

    题意：
        有n组数据，和怪兽血量m，每组数据有两个数，第一个为普通伤害值，
    第二个为怪兽血量少于该值时将造成双倍伤害，，求最少攻击次数，
    杀不死则输出-1.
    
    DFS：
        比较明显的dfs，时间复杂度为O(n!)，数据比较小而且不强，可以直接DFS
    过了 

*/
#include<stdio.h>
#include<string.h>
int n,m;
int a[10][2];
int vis[10];
int cnt;
void dfs(int c,int s)
{
    if(s<=0 && c<cnt){
        cnt=c;
        return;
    }
    if(c>=cnt) return;
    for(int i=0;i<n;i++)
        if(!vis[i]){
            vis[i]=1;
            int temp=(s<=a[i][1]?s-2*a[i][0]:s-a[i][0]);
            dfs(c+1,temp);
            vis[i]=0;
        }
        
}
int main(void)
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&a[i][0],&a[i][1]);
        memset(vis,0,sizeof(vis));
        cnt=10;
        dfs(0,m);
        if(cnt==10) puts("-1");
        else printf("%d
",cnt);
    }
}