POJ-3268(来回最短路+dijkstra算法)

Silver Cow Party

POJ-3268

• 这题也是最短路的模板题，只不过需要进行两次求解最短路，因为涉及到来回的最短路之和。
• 该题的求解关键是：求解B-A的最短路时，可以看做A是起点，这就和求解A-B的最短路很类似了，只不过需要将单向路的距离调换一下即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF = 1111111111;
int w[1001][1001];//d[i][j]表示i到j的距离
int d[1001];//d[i]表示s到i的最短近距离
int v[1001];//v[i]表示是否被访问过
int n;//表示顶点数
void dijkstra(int s) {
	for (int i = 0; i<n; i++) {
		if (i == s)
			d[i] = 0;
		else
			d[i] = INF;
	}
	memset(v, 0, sizeof(v));
	for (int i = 0; i<n; i++) {
		int x, m = INF;
		for (int y = 0; y<n; y++)
			if (!v[y] && d[y] <= m)
				m = d[x = y];
		v[x] = 1;
		for (int y = 0; y<n; y++)
			d[y] = min(d[y], d[x] + w[x][y]);
	}

}
int main() {
	int m, x;
	cin >> n >> m >> x;
	int a, b;
	int weight;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			w[i][j] = INF;
			if (i == j)
				w[i][j] = 0;
		}
	}
	for (int i = 0; i < m; i++) {
		cin >> a >> b>>weight;
		w[a-1][b-1] = weight;
	}
	dijkstra(x - 1);
	int to[1001];
	int back[1001];
	for (int i = 0; i < n; i++)
		back[i] = d[i];
	/*for (int i = 0; i < n; i++) {
		dijkstra(i);
		to[i] = d[x - 1];
	}*/
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < i; j++) {
			int temp = w[i][j];
			w[i][j] = w[j][i];
			w[j][i] = temp;
		}
	}
	dijkstra(x - 1);
	int max=0, max_i=0;

	for (int i = 0; i < n; i++) {
		if (max < d[i] + back[i]) {
			max = d[i] + back[i];
			max_i = i;
		}
	}
	cout << max << endl;
	return 0;
}

JAVA:

package POJ;
import java.util.*;
public class POJ_3268 {
	static int n,m,x;//n:1000,m:100000
	static final int INF = 1111111111;
	static int [][]w;//d[i][j]表示i到j的距离
	static int []d;//d[i]表示s到i的最短近距离
	static int []v;//v[i]表示是否被访问过
	static void dijkstra(int s) {
		for (int i = 0; i<n; i++) {
			if (i == s)
				d[i] = 0;
			else
				d[i] = INF;
		}
		Arrays.fill(v, 0);//这里很重要，因为v曾经改变过
		for (int i = 0; i<n; i++) {
			int x=0, m = INF;
			for (int y = 0; y<n; y++)
				if (v[y]==0 && d[y] <= m)
					m = d[x = y];
			v[x] = 1;
			for (int y = 0; y<n; y++)
				d[y] = Math.min(d[y], d[x] + w[x][y]);
		}

	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner cin=new Scanner(System.in);
		n=cin.nextInt();m=cin.nextInt();x=cin.nextInt();
		w=new int[n][n];
		d=new int[n];
		v=new int[n];
		int a, b;
		int weight;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				w[i][j] = INF;
				if (i == j)
					w[i][j] = 0;
			}
		}
		for (int i = 0; i < m; i++) {
			a=cin.nextInt();b=cin.nextInt();weight=cin.nextInt();
			w[a-1][b-1] = weight;
		}
		dijkstra(x - 1);
		int []to=new int[n];
		int []back=new int[n];
		for (int i = 0; i < n; i++)
			back[i] = d[i];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < i; j++) {
				int temp = w[i][j];
				w[i][j] = w[j][i];
				w[j][i] = temp;
			}
		}
		dijkstra(x - 1);
		int max=0, max_i=0;

		for (int i = 0; i < n; i++) {
			if (max < d[i] + back[i]) {
				max = d[i] + back[i];
				max_i = i;
			}
		}
		System.out.println(max);
	}

}