ACM-DFS之Beat——hdu2614

Beat
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.

依舊是DFS練習題，這道題講述的是一個孩子要做n個題目，每個題目難度不同，這個孩子有一個脾氣，就是做的題目不會做比之前做的簡單的。 題目中的Tij就是第i行第j列代表著，做完i號題目後做j題目所花費的時間，如果做i題目花費了2分鐘，做完i後做j花費時間小於2分鐘則，這個有脾氣的小屁孩不會去鳥這道題，他只做花費時間大於等於2分鐘的題目。 很有原則的小孩兒啊。 這個我主要想說的就是dfs函數中的flag，這個的設置比較巧妙。 代表每次最大值判斷（maxx和sl的比較）都是在解決了這道題後的， 防止循環到一道題，發現做完這道題後面的題目都不能做 或者 做完所有題目後的判斷。

代碼：

#include 
#include 
using namespace std;
int problem[16][16],vis[16];
int n,maxx;

void dfs(int line,int pre,int sl)
{
    int i,flag=0;
    for(i=0;i=pre)
        {
            vis[i]=1;
            dfs(i,problem[line][i],sl+1);
            vis[i]=0;
            flag=1;
        }
    }
    if(!flag)  maxx=sl>maxx?sl:maxx;
}

int main()
{
    int i,j;
    while(cin>>n)
    {
        memset(vis,0,sizeof(vis));
        for(i=0;i>problem[i][j];

        // 初始化maxx，開始dfs
        vis[0]=1;
        maxx=-1;
        dfs(0,0,1);
        cout<