HDU 2614 Beat (dfs)
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
題意：有n道題，a[i][j]表示做完i題後做j題的難度，這個人從第0題開始做，每次做不比上次題目簡單的題，問他最多做多少題？
理解題意半天，解題還是比較順利，解釋在代碼中
#include
#include
#include
#include
using namespace std;
#define N 20
int a[N][N],n;
int ans,vis[N];
void dfs(int le,int num,int s) //le表示剛做的題，例如le=0，那麼現在在第0行選擇做的下一題，s是剛做題的難度
{
int i;
if(num>ans) ans=num; //num是已經做的題
for(i=0;i
