UVa 11015 - 05-2 Rendezvous

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;
}