題目：有一個班級的學生要一起寫作業，所以他們要到一個統一的地點。現在給你他們各自的位置，

問集合地點定在哪，能够讓全部人走的總路徑長度最小。

分析：圖論、最短路。直接利用Floyd計算最短路，找到和值最小的輸出就可以。

說明：又是太長時間沒刷題了。╮(╯▽╰)╭。

#include <algorithm>

#include <iostream>

#include <string>

#include <map>

using namespace std;

int dist[23][23]; 

int main()

{

	int 	n, m, u, v, d, cases = 1;

	string  place;

	while (cin >> n >> m && n+m) {

		map<int, string>nameList;

		for (int i = 0; i < n; ++ i) {

			cin >> place;

			nameList[i] = place;

		}

		for (int i = 0; i < n; ++ i) {

			for (int j = 0; j < n; ++ j)

				dist[i][j] = 500000;

			dist[i][i] = 0;

		}

		for (int i = 0; i < m; ++ i) {

			cin >> u >> v >> d;

			dist[u-1][v-1] = dist[v-1][u-1] = d;

		}

		for (int k = 0; k < n; ++ k)

			for (int i = 0; i < n; ++ i)

				for (int j = 0; j < n; ++ j)

					if (dist[i][j] > dist[i][k]+dist[k][j])

						dist[i][j] = dist[i][k]+dist[k][j];

		int space = 0, max = 500000;

		for (int i = 0; i < n; ++ i) {

			int sum = 0;

			for (int j = 0; j < n; ++ j)

				if (i != j)

					sum += dist[i][j];

			if (max > sum) {

				max = sum;

				space = i;

			}

		}

		cout << "Case #" << cases ++ << " : " << nameList[space] << endl;

	}

    return 0;

}