POJ1861 Network(最小生成树)

题意：

有n个顶点，m条边，每条边双向的，并且有一定的长度，现在要求使每个顶点连通起来，并且使总长度最短输出最大边，和边的总数，还有是这些边的情况输出

要点：

这破题，special judge给的输出例子是错的，导致我先用prim算法搞了一个小时还是WA，怒换Kruskal算法10分钟AC了。不过用这题也算理解了special judge的意思，就是有很多种正确输出，只要选一种输出就行。后来发现我的prim算法模板有点问题，要先将map赋予一个比较大的值，否则比较最小值会出错，而且这题要求双向边，我之前没彻底理解prim。

Kruskal算法：

15346658

Seasonal

1861

Accepted

360K

63MS

C++

1063B

2016-04-03 10:04:01

#include
#include
#include
#include
#define maxn 1005
using namespace std;
int p[maxn],a[maxn],b[maxn];
int m, n,num;
struct edge
{
	int u, v, len;
}e[maxn*maxn];

bool cmp(edge a, edge b)
{
	return a.len < b.len;
}
void init()
{
	for (int i = 1; i <= m; i++)
		p[i] = i;
}
int find(int x)
{
	if (p[x] == x) return x;
	return p[x] = find(p[x]);
}
bool merge(int x, int y)
{
	x = find(x);
	y = find(y);
	if (x != y)
	{
		p[x] = y;
		return true;
	}
	return false;
}
int kruskal()
{
	init();
	sort(e, e + n, cmp);
	int max = -1, edges = 0;
	num = 0;
	for (int i = 0; i < n; i++)
	{
		if (merge(e[i].u, e[i].v))
		{
			if (max < e[i].len)
				max = e[i].len;
			a[num] = e[i].u; b[num++] = e[i].v;
			edges++;
		}
		if (edges + 1 == m)
			return max;
	}
	return -1;
}

int main()
{
	int x, y, temp;
	scanf("%d%d", &m, &n);
	for (int i = 0; i < n; i++)
		scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].len);
	printf("%d\n", kruskal());
	printf("%d\n", num);
	for (int i = 0; i < num; i++)
		printf("%d %d\n", a[i], b[i]);
	return 0;
}

Prim算法:

15346706

Seasonal

1861

Accepted

4124K

125MS

C++

1060B

2016-04-03 10:26:43

#include
#include
#include
int map[1005][1005], low[1005], pre[1005];
int m, n, num,max;
bool vis[1005];
int a[1005], b[1005];

void prim()
{
	int i, j, min, mark;
	num = 0; max = 0;
	memset(vis, false, sizeof(vis));
	for (i = 2; i <= m; i++)
	{
		low[i] = map[1][i];
		pre[i] = 1;//用pre数组记录前驱
	}
	vis[1] = true;
	for (i = 1; i < m; i++)
	{
		min = 0xffffff;
		for (j = 1; j <= m; j++)
			if (!vis[j] && low[j] < min)
			{
				min = low[j];
				mark = j;
			}
		if (max < min)
			max = min;
		vis[mark] = true;
		a[num] = pre[mark]; b[num++] = mark;
		for (j = 1; j <= m; j++)
			if (!vis[j] && map[mark][j] < low[j])
			{
				low[j] = map[mark][j];
				pre[j] = mark;//此时前驱更新为mark
			}
	}
}

int main()
{
	int x, y, temp;
	scanf("%d%d", &m, &n);
	for (int i = 1; i <= m; i++)
		for (int j = 1; j <= m; j++)
			map[i][j] = 0xfffff;//一开始要先赋予一个较大值
	for (int i = 0; i < n; i++)
	{
		scanf("%d%d%d", &x, &y, &temp);
		map[x][y] = temp;
		map[y][x] = temp;//要双向边
	}
	prim();
	printf("%d\n%d\n", max,num);
	for (int i = 0; i < num; i++)
		printf("%d %d\n", a[i], b[i]);
	return 0;
}