HDU 2614 Beat （DFS）

Beat

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1346    Accepted Submission(s): 793

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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题解：有一个人刷题，这个人准备刷n道题，但他刷的下一道题不会比前一道刷的题要简单，至少>=的难度。Orz.....先膜一下。。。

注意：题目中的Tij就是第i行第j列代表着，刷完 i 号题目后刷 j 题目所花费的时间，如果刷 i 题目花费了2分钟，刷完 i 后刷 j 花费时间小于2分钟，这个人不会去刷这道题，他只刷花费时间大于等于2分钟的题目。Orz。。。

        DFS。。。。

AC代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;int vis[21];int map[20][20];int sum; int n;void dfs(int k, int num,int time){for(int i=0;i<n;i++){if(vis[i]==0 &&map[k][i]>=time){vis[i]=1;dfs(i,num+1,map[k][i]);vis[i]=0;}}sum=max(sum,num);}int main(){ios::sync_with_stdio(false);while(cin>>n&&n){for(int i=0;i<n;i++){for(int j=0;j<n;j++){cin>>map[i][j];}}memset(vis,0,sizeof(vis));sum=0;vis[0]=1;dfs(0,1,0);cout<<sum<<"\n";}return 0;}