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.
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).
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
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1 //109MS 228K 694 B G++ 姜伯约 2 /* 3 4 题意： 5 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， 6 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， 7 杀不死则输出-1. 8 9 DFS： 10 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 11 过了 12 13 */ 14 #include<stdio.h> 15 #include<string.h> 16 int n,m; 17 int a[10][2]; 18 int vis[10]; 19 int cnt; 20 void dfs(int c,int s) 21 { 22 if(s<=0 && c<cnt){ 23 cnt=c; 24 return; 25 } 26 if(c>=cnt) return; 27 for(int i=0;i<n;i++) 28 if(!vis[i]){ 29 vis[i]=1; 30 int temp=(s<=a[i][1]?s-2*a[i][0]:s-a[i][0]); 31 dfs(c+1,temp); 32 vis[i]=0; 33 } 34 35 } 36 int main(void) 37 { 38 while(scanf("%d%d",&n,&m)!=EOF) 39 { 40 for(int i=0;i<n;i++) 41 scanf("%d%d",&a[i][0],&a[i][1]); 42 memset(vis,0,sizeof(vis)); 43 cnt=10; 44 dfs(0,m); 45 if(cnt==10) puts("-1"); 46 else printf("%d\n",cnt); 47 } 48 }
