poj-1125 Stockbroker Grapevine *

/*
 * 水题~   floyd算法 0ms
 *
 * 求每对顶点间的最短路， 然后选择 最大值 最小的那个顶点作为起点~
 */

#include <cstdio>
#include <cstring>
using namespace std;

const int maxN = 100 + 5;
const int inf = 10000000;
int n, totEdgeNum, w[maxN][maxN];

/*
struct SEdge{
    int u, v, w;
};
SEdge edge[maxN * maxN];
*/

void floyd(){
    for(int k=1; k<=n; k++){
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                if(w[i][j] > w[i][k] + w[k][j])
                    w[i][j] = w[i][k] + w[k][j];
            }
        }
    }
}

void getAns(){
    int totMin = inf, beg;
    for(int i=1; i<=n; i++){
        int lineMax = -1;
        bool flag = 1;
        for(int j=1; j<=n; j++){
            if(w[i][j] == -1){
                flag = 0; break;
            }
            if(lineMax < w[i][j]) lineMax = w[i][j];
        }
        if(flag && totMin > lineMax){
            totMin = lineMax; beg = i;
        }
    }

    if(totMin == inf)
        printf("disjoint\n");
    else
        printf("%d %d\n", beg, totMin);

}

int main(){
    while(scanf("%d", &n)){
        if(n == 0) return 0;

        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                w[i][j] = inf;

        int tmpN, tmpE, tmpW;
        for(int i=1; i<=n; i++){
            scanf("%d", &tmpN);
            for(int j=1; j<=tmpN; j++){
                scanf("%d%d", &tmpE, &tmpW);
                w[i][tmpE] = tmpW;
            }
            w[i][i] = 0;
        }

        floyd();

        getAns();
    }

    return 0;
}