题意：给n个节点m条边，给出每条边相连节点，与边的权值，求1到n的最短路
Dijkstra算法：

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm> 
#include<cstring>
using namespace std;
int n, m, g[105][105], path[105], vis[105];
int main()
{
	while (cin >> n >> m && n && m) {
		memset(g, 0x3f3f3f3f, sizeof(g));
		memset(vis, 0, sizeof(vis));
		memset(path, 0x3f3f3f3f, sizeof(path));
		for (int i = 0; i < m; i++) {
			int x, y, dis;
			cin >> x >> y >> dis;
			g[x][y] = min(dis, g[x][y]);
			g[y][x] = g[x][y];
		}
		for (int i = 1; i <= n; i++) {
			path[i] = g[1][i];
		}
		while (true) {
			int k = -1;
			for (int i = 1; i <= n; i++)
				if (!vis[i] && (k == -1 || path[i] < path[k]))
					k = i;
			if (k == -1)
				break;
			vis[k] = 1;
			for (int i = 1; i <= n; i++)
				if (!vis[i] && path[k] + g[k][i] < path[i]) {
					path[i] = path[k] + g[k][i];
				}
		}
		cout << path[n] << endl;
	}
}

Floyd算法：

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm> 
#include<cstring>
using namespace std;
long long n, m, g[105][105];
int main()
{
	while (cin >> n >> m && n && m) {
		memset(g, 0x3f3f3f3f, sizeof(g));
		for (int i = 0; i < m; i++) {
			long long x, y, dis;
			cin >> x >> y >> dis;
			g[x][y] = min(g[x][y], dis);
			g[y][x] = g[x][y];
		}
		for (int k = 1; k <= n; k++)
			for (int i = 1; i <= n; i++)
				for (int j = 1; j <= n; j++)
					if (g[i][j] > g[i][k] + g[k][j]) {
						g[i][j] = g[i][k] + g[k][j];
					}
		cout << g[1][n] << endl;
	}
}