UVA - 567 Risk

题意：求图上两点的最短路程，Floyd的模板题，输入的时候是按1-19对应的连接点输入的

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 30;int d[MAXN][MAXN],N,cas;void init(){    memset(d,0x3f3f3f3f,sizeof(d));    for (int i = 1; i <= 20; i++)        d[i][i] = 0;}bool read(){    init();    int n,a;    if (scanf("%d",&n) == EOF)        return 0;    for (int i = 0; i < n; i++){        scanf("%d",&a);        d[1][a] = 1;        d[a][1] = 1;    }    for (int i = 2; i <= 19; i++){        scanf("%d",&n);        for (int j = 0; j < n; j++){            scanf("%d",&a);            d[i][a] = 1;            d[a][i] = 1;        }    }    return 1;}void Floyd(){    for (int k = 1; k <= 20; k++)        for (int i = 1; i <= 20; i++)            for (int j = 1; j <= 20; j++)                d[i][j] = min(d[i][j],d[i][k]+d[k][j]);}void solve(){    printf("Test Set #%d\n",cas++);    scanf("%d",&N);    int u,v;    while (N--){        scanf("%d%d",&u,&v);        printf("%2d to %2d: %d\n",u,v,d[u][v]);    }    printf("\n");}int main(){    cas = 1;    while (read()){        Floyd();        solve();    }}