POJ 1789 Truck History

/*
*POJ 1789 Truck History
*不同类型卡车之间的距离为编码对应位不同的个数(如 aaaaaaa 和 babaaaa 距离为2)
*求最小生成树
*/

#include <cstdio>
#include <cstring>
#define INF 1000000
#define MAXN 2000
#define CODELEN 7
int n;
char codes[MAXN][CODELEN + 1]; //每种卡车编码
int d[MAXN][MAXN]; //邻接矩阵
int lowcost[MAXN]; // 节点i 距离已合并集合的最小权值

void init()
{
	int i, j, k;
	int dist;
	memset(d, 0, sizeof(d));
	for (i = 0; i < n; i++) {
		for (j = i + 1; j < n; j++) {
			dist = 0;
			for (k = 0; k < 7; k++) {
				dist += codes[i][k] != codes[j][k];
			}
			d[i][j] = d[j][i] = dist;
		}
	}
}

int prim()
{
	int i, j, k;
	int sum = 0;
	lowcost[0] = -1; //从顶点0开始构造
	for (i = 1; i < n; i++) {
		lowcost[i] = d[0][i];
	}
	for (i = 1; i < n; i++) {
		int min = INF;
		for (k = 0; k < n; k ++) {
			if (lowcost[k] != -1 && lowcost[k] < min) {
				j = k;
				min = lowcost[k];
			}
		}
		sum += min;
		lowcost[j] = -1; //把顶点j加入生成树
		for (k = 0; k < n; k++) {
			if (d[j][k] < lowcost[k]) {
				lowcost[k] = d[j][k];
			}
		}
	}
	return sum;
}

int main()
{
	int i;
	while (scanf("%d", &n) && n) {
		for (i = 0; i < n; i++) {
			scanf("%s", codes[i]);
		}
		init();
		printf("The highest possible quality is 1/%d.\n", prim());
	}
	return 0;
}