UVA 125 Numbering Paths

题意 给出方向（有向）然后判断从一个点到另一个点的方案数。如果有无数条那么对应位置置为-1

直接先dp处理出来。dp[i][j] = sum(dp[i][k]*dp[k][j])  

同时如果两点之间有无限条路径。那么这两点之间必然有一环存在。有f[k][k]!=0 

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
#define MAXN 35
int dp[MAXN][MAXN];
int N,M;
void read()
{
    N = 0;
    memset(dp,0,sizeof(dp));
    while (M--)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        dp[u][v] = 1;
        N = max(N,max(u,v));
    }
}
int main()
{
    //freopen("sample.txt","r",stdin);
    int kase = 0;
    while (scanf("%d",&M) != EOF)
    {
        read();
        for (int k = 0; k <= N; k++)
            for (int i = 0; i <= N; i++)
             for (int j = 0 ; j <= N; j++)
             dp[i][j] += dp[i][k] * dp[k][j];
        for (int i = 0; i <= N; i++)
            if (dp[i][i])
        {
            for (int j = 0; j <= N; j++)
                for (int k = 0; k <= N; k++)
                if (dp[j][i] && dp[i][k]) dp[j][k] = -1;
        }
        printf("matrix for city %d\n",kase++);
        for (int i = 0; i <= N; i++)
        {
            for (int j = 0; j <= N; j++)
            {
                if (!j) printf("%d",dp[i][j]);
                else printf(" %d",dp[i][j]);
            }
            putchar('\n');
        }
    }
    return 0;
}