HDOJ 题目2614 Beat（DFS）

Beat

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 837    Accepted Submission(s): 524

Problem Description

Zty is a man that always full of enthusiasm. He wants to solve every kind of difficulty ACM problem in the world. And he has a habit that he does not like to solve
a problem that is easy than problem he had solved. Now yifenfei give him n difficulty problems, and tell him their relative time to solve it after solving the other one.
You should help zty to find a order of solving problems to solve more difficulty problem. 
You may sure zty first solve the problem 0 by costing 0 minute. Zty always choose cost more or equal time’s problem to solve.

 

Input

The input contains multiple test cases.
Each test case include, first one integer n ( 2< n < 15).express the number of problem.
Than n lines, each line include n integer Tij ( 0<=Tij<10), the i’s row and j’s col integer Tij express after solving the problem i, will cost Tij minute to solve the problem j.

 

Output

For each test case output the maximum number of problem zty can solved.

 

Sample Input

30 0 01 0 11 0 030 2 21 0 11 1 050 1 2 3 10 0 2 3 10 0 0 3 10 0 0 0 20 0 0 0 0

 

Sample Output

324
Hint
Hint: sample one, as we know zty always solve problem 0 by costing 0 minute. So after solving problem 0, he can choose problem 1 and problem 2, because T01 >=0 and T02>=0. But if zty chooses to solve problem 1, he can not solve problem 2, because T12 < T01. So zty can choose solve the problem 2 second, than solve the problem 1.  
 

 

Author

yifenfei

 

Source

奋斗的年代

 

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 ac代码

#include<stdio.h>#include<string.h>#define max(a,b) (a>b?a:b)int map[1010][1010],vis[1010];int ans,n;void dfs(int i,int len,int num){ans=max(ans,len);if(len==n)return;for(int j=0;j<n;j++){if(!vis[j]&&j!=i){if(map[i][j]>=num){vis[j]=1;dfs(j,len+1,map[i][j]);vis[j]=0;}}}}int main(){//int n;while(scanf("%d",&n)!=EOF){int i,j;for(i=0;i<n;i++){for(j=0;j<n;j++)scanf("%d",&map[i][j]);}memset(vis,0,sizeof(vis));vis[0]=1;ans=-1;for(i=1;i<n;i++){vis[i]=1;dfs(i,2,map[0][i]);vis[i]=0;}printf("%d\n",ans);}}