hdu2614

Beat

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1769    Accepted Submission(s): 1042

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

32
4
题意分析：有n道题目，a[i][j] 表示做完第 i 道题再做第 j 道题所要花费的时间。Zty每次只做比以前做过的题目更难的题目，也就是说时间比以前的长。问最多能做多少道题。(每次都从第0题开始做，第0题花费时间为0).简单深度搜索
AC代码：
#include<stdio.h>#include<stdlib.h>#include<math.h>#include<queue>#include<string.h>#include<iostream>#include<vector>using namespace std;int map[20][20];int vis[20]={0};int n;int maxn;void DFS(int k,int a,int num){        if(maxn==n)                return;        maxn=max(maxn,num);        for(int i=0;i<n;i++)        {                if(!vis[i]&&a<=map[k][i])                {                        vis[i]=1;                        DFS(i,map[k][i],num+1);                        vis[i]=0;                }        }}int main(){        while(scanf("%d",&n)!=EOF)        {                for(int i=0;i<n;i++)                {                        for(int j=0;j<n;j++)                                scanf("%d",&map[i][j]);                }                maxn=0;                vis[0]=1;                DFS(0,0,1);                vis[0]=0;                printf("%d\n",maxn);        }        return 0;}