题意:有几个村庄,要修最短的路,使得这几个村庄连通。但是现在已经有了几条路,求在已有路径上还要修至少多长的路。

分析:用Prim求最小生成树,将已有路径的长度置为0,由于0是最小的长度,所以一定会被Prim选中加入最小生成树。

package Map;

import java.util.Scanner;

/**
* Prime
*/
public class Poj_2421_Prim { static int MAXVEX = 200;
static int n, m;
static int[][] arc = new int[MAXVEX][MAXVEX];
static int visited[] = new int[MAXVEX];//判断是否加入生成树 public static int prime() { int min, i, j, k, sum = 0;
visited[1] = 1; for (i = 2; i <= n; i++) {
min = 1000000;
k = 0;
for (j = 1; j <= n; j++) {
if (visited[j] == 0 && arc[1][j] < min) {
min = arc[1][j];
k = j;
}
} sum += min;
visited[k] = 1;
for (j = 1; j <= n; j++) {
if (visited[j] == 0 && arc[1][j] > arc[k][j]) {
arc[1][j] = arc[k][j];
}
}
}
return sum;
} public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
n = sc.nextInt(); for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
arc[i][j] = sc.nextInt();
}
arc[i][i] = 1000000;
} m = sc.nextInt(); //如果路径存在,则置为0.这样
for (int i = 1; i <= m; i++) {
int s = sc.nextInt();
int e = sc.nextInt();
arc[s][e] = 0;
arc[e][s] = 0;
} System.out.println(prime());
}
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

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