ZOJ - 1203 Swordfish （非負權值的最小生成樹/最短路 - Kruskal算法）

題目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=203

題意：

將n個城市，全部連通起來的最短長度

分析：

n個點，每個點與其他n-1個點均可相連，距離d由座標計算可得

兩點距離爲每條邊的權值，直接Kruskal算了得到最小生成樹結果

核心：

void Kruskal()
{
	int ans = 0, num = 0;
	for(i = 1; i<m; i++)
	{
		if(Find( x[i] ) != Find( y[i] ))
		{
		ans += w[	i];
		num++;
		Union( x[i], y[i] );
		}
		if(num >= n-1) break;
	}
	printf("%	d\n", ans);
}

代碼：

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

using namespace std;

#define MAXN 110
#define MAXM 5010
#define MIN -1e+10
#define INF 0x7f7f7f7f



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

int n, m, i, j;
double x[MAXN], y[MAXN];
double ans;
int parent[MAXN];
int Case;

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 x, int y)
{
	int r1 = Find(x), r2 = Find(y);
	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));
	ans = 0;
	int num = 0;
	int 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;
	}
	if(Case>1)printf("\n");
	printf("Case #%d:\n", Case++);
	printf("The minimal distance is: %.2lf\n", ans);
}



int main()
{
	//freopen("a.txt", "r", stdin);
	
	Case = 1;
	while(~scanf("%d", &n) && n)
	{
		for(i = 0; i<n; i++)
		{
			scanf("%lf%lf", &x[i], &y[i]);
		}
		double d;
		m = 0;
		for(i = 0; i<n; i++)
		{
			for(j = i+1; j<n; j++)
			{
				d = sqrt((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
				edges[m].u = i;
				edges[m].v = j;
				edges[m].w = d;
				m++;
			}
		}
		sort(edges, edges + m);
		Kruskal();
	}
	return 0;
}