Beat Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 349    Accepted Submission(s): 225


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
3 0 0 0 1 0 1 1 0 0 3 0 2 2 1 0 1 1 1 0 5 0 1 2 3 1 0 0 2 3 1 0 0 0 3 1 0 0 0 0 2 0 0 0 0 0
 

Sample Output
3 2 4 
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
 

Recommend
yifenfei


解题思路：明显的DFS题目，只不过，注意顺序，做完题目i后,可以按照第i行的数据选择下一个要做的题目，数据量不大，不用剪枝，直深搜就好了。


#include<stdio.h> #include<cstring> using namespace std; int map[15][15]; int flag[16];     //做题情况记录，做过的题目就不用再做了 int sum,n; void dfs(int max2,int sum1,int j) //用于验证的值,当前解题数量,当前行 {     //printf("dfs\n");     for(int i=0;i<n;i++)     {         if(!flag[i]&&map[j][i]>=max2)   //不大于的意思就是，小于或等于//做过的题目就不用再做了         {             flag[i]=1;             sum=sum>sum1+1?sum:sum1+1;             //printf("max2=%d map[i][j]=%d sum1+1=%d j=%d  f[%d]=1\n",max2,map[j][i],sum1+1,j,i);             dfs(map[j][i],sum1+1,i);    //做了第i题了，后面按照第i行的数据选题             flag[i]=0;         }     } } int main() {     int i,j;     int t;     while(scanf("%d",&n)!=EOF)     {         sum=0;         for(i=0;i<n;i++)             for(j=0;j<n;j++)                 scanf("%d",&map[i][j]);         memset(flag,0,sizeof(flag));         flag[0]=1;         dfs(0,1,0);         printf("%d\n",sum);     }     return 0; }