POJ2421 Constructing Roads(最小生成树)

题意：

村庄之间连通，要求输出最小权值和，注意有些村庄已经连通

要点：

Prim算法稍微变一下，因为是求权值和，已经连通的之间权值改为0即可，反正为0后面算的时候肯定是最小值，会加入集合的，然后权值又是0不影响。

15347342

Seasonal

2421

Accepted

208K

32MS

C++

826B

2016-04-03 14:36:19

#include
#include
#include
int map[105][105], low[105];
bool vis[105];
int n, m;

int prim()
{
	int i, j,mark,min,sum=0;
	memset(vis, false, sizeof(vis));
	for (i = 1; i <= n; i++)
		low[i] = map[1][i];
	vis[1] = true;
	for (i = 2; i <= n; i++)
	{
		min = 0xfffff;
		for (j = 1; j <= n;j++)
			if (!vis[j] && min > low[j])
			{
				min = low[j];
				mark = j;
			}
		vis[mark] = true;
		sum += min;
		for (j = 1; j <= n; j++)
			if (!vis[j] && map[mark][j] < low[j])
				low[j] = map[mark][j];
	}
	return sum;
}
int main()
{
	int i, j,x,y;
	while (~scanf("%d", &n))
	{
		for (i = 1; i <= n; i++)
			for (j = 1; j <= n; j++)
				scanf("%d", &map[i][j]);
		scanf("%d", &m);
		while (m--)
		{
			scanf("%d%d", &x, &y);
			map[x][y] = 0;		//这两个已经通了，那么直接赋值为0，加起来不影响结果
			map[y][x] = 0;
		}
		printf("%d\n", prim());
	}
	return 0;
}

转载于:https://www.cnblogs.com/seasonal/p/10343796.html