您的位置:首页 > 其它

Kruskal 最小生成树(输出最小生成树的路径,权值的和) 参考poj 1258

2012-06-15 17:39 405 查看
输入无向图的,矩阵存储格式,

package com.liang.poj;

import java.util.Scanner;

public class Test2485Kruskal {
static int n = 0;
static int[][] cin = null;
static boolean[] visited = null;        //对顶点进行标记
static int[][] brimCollection = null;   //边的集合
static int index = 0;
static int[] path = null;              //记录路径和路径边上的权值
static int pathIndex = 0;

public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
n = scan.nextInt();
cin = new int

;
visited = new boolean
;
brimCollection = new int[n * n - n][3];
path = new int[n - 1];

for (int i = 0; i < cin.length; i++) { // 初始化cin brimCollection
for (int j = 0; j < cin[i].length; j++) {
cin[i][j] = scan.nextInt();
if (cin[i][j] != 0) {
brimCollection[index][0] = i;
brimCollection[index][1] = j;
brimCollection[index++][2] = cin[i][j];
}
}
}

sort();

path[pathIndex++] = 0;
visited[brimCollection[0][0]] = true;
visited[brimCollection[0][1]] = true;

kruskal();

for(int i=0; i<path.length; i++) {                       //对路径和权值进行输出
System.out.print(brimCollection[path[i]][0] + " ");
System.out.print(brimCollection[path[i]][1] + " ");
System.out.println();
}

int sum = 0;
for(int i=0; i<path.length; i++) {
sum += brimCollection[path[i]][2];
}
System.out.println(sum);       //输出最小生成树,权值的和

}
}

public static void kruskal() {      //克鲁斯卡尔算法
int count = 1;
for (int i = 1; i < brimCollection.length; i++) {
int a = brimCollection[i][0];
int b = brimCollection[i][1];
if ((!visited[a] && visited[b]) || (visited[a] && !visited[b])) {
visited[a] = true;
visited[b] = true;
path[pathIndex++] = i;
i = 0;
count ++ ;
}
if(count == n) {
break;
}
}
}

public static void sort() {            //对所有的边按权值进行排序
for (int i = 0; i < brimCollection.length; i++) {
if (brimCollection[i][2] == 0) {
break;
}
for (int j = i + 1; j < brimCollection.length; j++) {
if (brimCollection[j][2] == 0) {
break;
}
if (brimCollection[i][2] > brimCollection[j][2]) {
exchange(i, j, 0);
exchange(i, j, 1);
exchange(i, j, 2);
}
}
}
}

public static void exchange(int i, int j, int cos) {
int x = 0;
x = brimCollection[i][cos];
brimCollection[i][cos] = brimCollection[j][cos];
brimCollection[j][cos] = x;
}

}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: