UVA - 567 Risk(Floyd）

题目链接

题目大意：有20个城市，输入给19行，每行先给有几个数，然后接着给出这几个数，代表的是后面的城市编号和行编号（城市编号）有一条边，每条边的权值为1.接着m个查询任意两个城市之间的最短距离。

解题思路：求任意两个顶点之间的距离，用floyd。

代码:

#include <cstdio>

const int maxn = 21;
const int INF = 0x3f3f3f3f;

int G[maxn][maxn];

void Floyd () {

    for (int k = 1; k < maxn; k++)
        for (int i = 1; i < maxn; i++)
            for (int j = 1; j < maxn; j++)
                if (G[i][j] > G[i][k] + G[k][j])
                    G[i][j] = G[j][i] = G[i][k] + G[k][j];
}

int main () {

    int n, m, u, v, cas = 0;

    while (scanf ("%d", &n) != EOF) {

        for (int i = 1; i < maxn; i++)
            for (int j = 1; j < maxn; j++)
                G[i][j] = (i == j) ? 0 : INF;

        for (int i = 0; i < n; i++) {
            scanf ("%d", &v);
            G[1][v] = G[v][1] = 1;
        }

        for (int i = 2; i < 20; i++) {
            scanf ("%d", &n);
            for (int j = 0; j < n; j++) {
                scanf ("%d", &v);
                G[i][v] = G[v][i] = 1;
            }
        }   

        Floyd();

        scanf ("%d", &m);
        printf ("Test Set #%d\n", ++cas);
        while (m--) {

            scanf ("%d%d", &u, &v);
            printf ("%2d to %2d: %d\n", u, v, G[u][v]);
        }
        printf ("\n");

    }
    return 0;
}