POJ - 1251 Jungle Roads （最小生成樹Kruskal、並查集）

題目：http://poj.org/problem?id=1251

題意：

n個村子，每個村子後面連接各自的鄰村、路費，選出使得全部村子相連的最小通路

分析：

n個節點，村子之間的路費爲邊的權值，Kruskal算法求最小生成樹

核心：

void Kruskal()
{
	memset(parent, -1, sizeof(parent));
	int num = 0, ans = 0, u, v;
	for(i = 0; i<m; i++)
	{
		u = edges[i].u; v = edges[i].v;
		if(Find(u) != Find(v))
		{
			ans += edges[i].w;
			num++;
			Union(u, v);
		}
		if(num >= n-1) break;
	}
	printf("%d\n", ans);
}

代碼：

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>

using namespace std;

#define MAXN 30
#define MAXM 100
#define MIN -1e+10
#define INF 0x7f7f7f7f

struct edge
{
	int u, v, w;
	bool operator < (const edge& p)const
	{
		return w<p.w;
	}
}edges[MAXM];

int parent[MAXN];
int n, m, i, j;

int Find(int x)
{
	int s, temp;
	for(s = x; parent[s]>=0; s = parent[s]);
	while(s != x)
	{
		temp = parent[x];
		parent[x] = s;
		x = temp;
	}
	return s;
}

void Union(int u, int v)
{
	int r1 = Find(u), r2 = Find(v);
	int temp = parent[r1] + parent[r2];
	if(parent[r1] > parent[r2])
	{
		parent[r1] = r2;
		parent[r2] = temp;
	}
	else 
	{
		parent[r2] = r1;
		parent[r1] = temp;
	}
}

void Kruskal()
{
	memset(parent, -1, sizeof(parent));
	int num = 0, ans = 0, u, v;
	for(i = 0; i<m; i++)
	{
		u = edges[i].u; v = edges[i].v;
		if(Find(u) != Find(v))
		{
			ans += edges[i].w;
			num++;
			Union(u, v);
		}
		if(num >= n-1) break;
	}
	printf("%d\n", ans);
}

int main()
{
	freopen("a.txt", "r", stdin);

	char s[2];
	int nt, w, u, v;
	while(~scanf("%d", &n) && n)
	{
		m = 0;
		for(i = 1; i<n; i++)
		{
			scanf("%s%d", s, &nt);
			u = s[0] - 'A' + 1;
			for(j = 1; j<=nt; j++)
			{
				scanf("%s%d", s, &w);
				edges[m].u = u;
				edges[m].v = s[0] - 'A' + 1;
				edges[m].w = w;
				m++;
			}
		}
		sort(edges, edges+m);
		Kruskal();
	}
	return 0;
}