九度1008,最短路径问题

AC

import java.util.Arrays;
import java.util.Scanner;

public class J1008_2 {

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		int n = cin.nextInt();
		int m = cin.nextInt();
		while (n != 0 && m != 0) {
			int[][] map = new int[n + 5][n + 5];
			for (int i = 0; i < n + 5; i++)
				Arrays.fill(map[i], Integer.MAX_VALUE / 3);
			int[][] w = new int[n + 5][n + 5];
			for (int i = 0; i < n + 5; i++)
				Arrays.fill(w[i], Integer.MAX_VALUE / 3);

			for (int i = 1; i <= m; i++) {
				int a = cin.nextInt();
				int b = cin.nextInt();
				int c = cin.nextInt();
				int d = cin.nextInt();
				map[a][b] = c;
				map[b][a] = c;
				w[a][b] = d;
				w[b][a] = d;
			}
			int s = cin.nextInt();
			int t = cin.nextInt();
			int[] d = new int[n + 5];
			Arrays.fill(d, Integer.MAX_VALUE);
			int[] co = new int[n + 5];
			Arrays.fill(co, Integer.MAX_VALUE);
			boolean[] fa = new boolean[n + 5];
			Arrays.fill(fa, true);
			fa[s] = false;
			for (int i = 1; i <= n; i++) {
				d[i] = map[s][i];
				co[i] = w[s][i];
			}
			for (int i = 1; i < n; i++) {
				int min = Integer.MAX_VALUE;
				int k = 0;
				for (int j = 1; j <= n; j++) {
					if (fa[j] && (min > d[j] || (min == d[j] && co[k] > co[j]))) {
						min = d[j];
						k = j;

					}

				}
				fa[k] = false;
				for (int j = 1; j <= n; j++) {
					if (fa[j]
							&& (d[k] + map[k][j] < d[j] || (d[k] + map[k][j] == d[j] && co[k]
									+ w[k][j] < co[j]))) {
						d[j] = d[k] + map[k][j];
						co[j] = co[k] + w[k][j];
					}
				}

			}
			System.out.println(d[t] + " " + co[t]);

			n = cin.nextInt();
			m = cin.nextInt();
		}

	}

}

之前用了佛洛依德超时了，改成C也超时了，于是用了Dijkstra，Dijkstra和Prim很像，Dijkstra是通过节点k来更新d[j],即:d[j] = d[k] + map[k][j];

而Prim是用刚加入最小生成树的节点k到j的距离来更新节点j到原最小生成树的距离，即d[j]=map[k][j]