AOJ0189

题意

求某一办公室到其他办公室的最短距离。 
多组输入，n表示n条关系，下面n次每次输入 x y d表示x到y的距离是d。需要注意的是n没有给定，需要根据输入来求。 
输出办公室的编号和距离。

思路

任意两点之间的最短距离用floyd算法比较合适。

代码

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int N = 10;const int M = 10;const int INF = 0x3f3f3f3f;int d[N][N];int n;void solve(){    for (int k = 0; k < n; k ++) {        for (int i = 0; i < n; i ++) {            for (int j = 0; j < n; j ++) {                d[i][j] = min(d[i][j], d[i][k]+d[k][j]);            }        }    }    int sum, msum = INF;    int id = 0;    for (int i = 0; i < n; i ++) {        sum = 0;        for (int j = 0; j < n; j ++) {            if (i == j) continue;            sum += d[i][j];        }        if (sum < msum) {            msum = sum;            id = i;        }    }    printf("%d %d\n", id, msum);}int main(void){    int m;    while ( cin >> m && m ) {        for (int i = 0; i < N; i ++)            fill(d[i], d[i]+N, INF);        n = 0;        while ( m-- ) {            int a, b, c;            scanf("%d%d%d", &a, &b, &c);            d[a][b] = d[b][a] = c;            n = max(max(n, a+1), b+1);        }        solve();    }    return 0;}