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[][];
 int vis[];
 int cnt;
 void dfs(int c,int s)
 {
     if(s<= && c<cnt){
         cnt=c;
         return;
     }
     if(c>=cnt) return;
     for(int i=;i<n;i++)
         if(!vis[i]){
             vis[i]=;
             int temp=(s<=a[i][]?s-*a[i][]:s-a[i][]);
             dfs(c+,temp);
             vis[i]=;
         }

 }
 int main(void)
 {
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(int i=;i<n;i++)
             scanf("%d%d",&a[i][],&a[i][]);
         memset(vis,,sizeof(vis));
         cnt=;
         dfs(,m);
         if(cnt==) puts("-1");
         else printf("%d\n",cnt);
     }
 }