POJ 1502 MPI Maelstrom 最短路

题意：有n台主机，给出每两台主机之间的通信时间，每台主机允许向所有连接到它的主机同时发送消息，求一条消息从主机1发送到所有其他主机所需要的最少时间。
思路：题意给的不太清楚，乍一看还以为要求最小生成树，实际上就是求主机1到其他主机的最短路中的最大值，用djkstra求就行

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<stack> 
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e2+10;
const int inf = 0x3f3f3f3f;
int n, g[maxn][maxn], d[maxn], vis[maxn];
int main()
{
	while (cin >> n) {
		getchar();
		for (int i = 0; i < n; i++) {
			g[i][i] = inf;
			for (int j = 0; j < i; j++) {
				string s;
				cin >> s;
				if (s[0] == 'x')
					g[j][i] = g[i][j] = inf;
				else {
					int num = 0, cur = 0;
					for (int i = s.length()-1; i >= 0; i--) {
						num += (s[i]-'0')*pow(10, cur++);
					}
					g[j][i] = g[i][j] = num;
				}
			}
		}
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < n; i++)
			d[i] = inf;
		d[0] = 0;
		while (true) {
			int k = -1;
			for (int i = 0; i < n; i++)
				if (!vis[i] && (k == -1 || d[i] < d[k]))
					k = i;
			if (k == -1)
				break;
			vis[k] = 1;
			for (int i = 0; i < n; i++)
				if (d[i] > d[k] + g[k][i])
					d[i] = d[k] + g[k][i];
		}
		int ans = 0;
		for (int i = 1; i < n; i++)
			ans = max(ans, d[i]);
		cout << ans << "\n";
	} 
	return 0;
}