POJ 3241 曼哈顿最小生成树

//题意： 询问平面上的点的曼哈顿距离最小生成树第n-k小边的长度，点数为10000。
//最小生成树边权对应？？？
//1A2s险过
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#define inf 0x3f3f3f3f
#define maxn 10100
#define maxm 100000
using namespace std;
int n,k;
int cnt;

struct point
{
int x,y;
}G[maxn];

int pre[maxn],rank[maxn];

struct Edge
{
int l,r;
int w;
};

Edge edge[maxm];

void build(int target)
{
int dis;
int tempx=G[target].x;
int tempy=G[target].y;
int indexup=0;int indexdown=0;
int miup=inf;
int midown=inf;
for(int i=1;i<=n;i++)
{
if(i==target) continue;
dis=abs(G[i].x-tempx)+abs(G[i].y-tempy);
if(G[i].x>=tempx && G[i].y>tempy)
{
if(dis<miup  )
indexup=i,miup=dis;
}
if(G[i].x>tempx && G[i].y<=tempy )
{
if(dis<midown)
{
indexdown=i,midown=dis;
}
}
}
if(indexup!=0)
{
edge[cnt].l=indexup;///加边求最小生成树
edge[cnt].r=target;
dis=abs(G[indexup].x-tempx)+abs(G[indexup].y-tempy);
edge[cnt++].w=dis;
}
if(indexdown!=0)
{
dis=abs(G[indexdown].x-tempx)+abs(G[indexdown].y-tempy);
edge[cnt].l=indexdown;
edge[cnt].r=target;
edge[cnt++].w=dis;
}
}

void Init()
{
for(int i=0;i<=cnt+10;i++)
{
pre[i] = i;
rank[i] = 1;
}
}

int Find(int x)
{
if(pre[x] != x)
pre[x] = Find(pre[x]);
return pre[x];
}

void Union(int x,int y)//这是传说中按秩序并查么
{
x = Find(x);
y = Find(y);
if(x == y) return;
if(rank[x] >= rank[y])
{
pre[y] =x;
rank[x] += rank[y];
}
else
{
pre[x] = y;
rank[y] += rank[x];
}
}

bool cmp(Edge a,Edge b)
{
return a.w < b.w;
}

void Kruskal()
{
Init();
int ans = 0;
sort(edge,edge+cnt,cmp);
int edgenum=0;
for(int i=0;i<cnt;i++)
{
if(edgenum==n-1)
break;
int x = Find(edge[i].l);
int y = Find(edge[i].r);
if(x != y)
{
Union(x,y);
ans += edge[i].w;
edgenum++;
if(edgenum==n-k)
{
cout<<edge[i].w<<endl;return;
}

}
}

}

int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
cnt=0;
for(int i=1;i<=n;i++)
scanf("%d%d",&G[i].x,&G[i].y);
for(int i=1;i<=n;i++)
build(i);
Kruskal();
}
}