Beat

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

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.  
 

#include<stdio.h>

const int N = 1<<14;
const int inf = 0xffffff;

int k[N];
void init()
{
    for(int i=1;i<N; i++)
    {
        int n=0;
        for(int s=0;(1<<s)<=i; s++)
            if((1<<s)&i)
            n++;
        k[i]=n;
    }
}
int main()
{
    int n,map[14][14],dp[N][14],maxlen;
    init();
    while(scanf("%d",&n)>0)
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            scanf("%d",&map[i][j]);
        for(int s=1;s<(1<<n);s++)
        for(int i=0;i<n;i++)
            dp[s][i]=inf;

        dp[1<<0][0]=0;
        maxlen=0;

        for(int s=1; s<(1<<n); s++)
        for(int i=0; (1<<i)<=s; i++)
        if(dp[s][i]!=inf)
        {
            if(maxlen<k[s])maxlen=k[s];
            for(int j=0;j<n;j++)
                if(!((1<<j)&s)&&map[i][j]>=dp[s][i]&&dp[s|(1<<j)][j]>map[i][j])
                    dp[s|(1<<j)][j]=map[i][j];
        }
        printf("%d\n",maxlen);
    }
}

HDU2614Beat(DP状态压缩路径)

原文：http://blog.csdn.net/u010372095/article/details/41849705