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.
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
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; } |
