poj 1258 prim算法求最小生成树

 

#include <stdio.h>
#include <string.h>

#define DEBUG

#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__) 
#else
#define debug(...)
#endif

#define N 100
#define MAX_INT 20000000

int 	graph[N][N];
int 	dist[N];	/* dist[i]表示顶点i距离已加入集合的最小距离, 如果dist[i]变为-1, 表示顶点i已加入集合 */

int n;

int prim(int u)
{
	int 	i, p, min, total_cost, min_edge;

	for (i = 0; i < n; i++) {
		dist[i] = MAX_INT;
	}

	p = u;	/* p为选种加入集合的点,初始时为u */
	total_cost = 0;
	for (;;) {
		min = MAX_INT;
		min_edge = -1;
		debug("顶点%d加入集合\n", p);
		dist[p] = -1;
		/* 把顶点p加入集合后, 遍历所有未加入集合的顶点, 更新它们的距离, 同时找出距离最小的顶点 */
		for (i = 0; i < n; i++) {
			if (dist[i] >= 0) {		/* 顶点i未加入集合 */
				if (graph[p][i] > 0 && graph[p][i] < dist[i]) {
					dist[i] = graph[p][i];
					debug("更新顶点%d的距离为%d\n", i, dist[i]);
				}
				if (dist[i] < min) {
					min = dist[i];
					min_edge = i;
				}
			}	
		}
		if (min_edge == -1) break;
		debug("距离最小的顶点为%d, 长度为%d\n", min_edge, min);
		p = min_edge;
		total_cost += min;
	}
	return total_cost;
}

int main()
{
	int 	i, j;

	while (scanf("%d", &n) != EOF) {
		for (i = 0; i < n; i++) {
			for (j = 0; j < n; j++) {
				scanf("%d", &graph[i][j]);
			}
		}
		printf("%d\n", prim(0));
	}
	return 0;
}