Thinking minimum spanning tree + diffusion (Los Valley P1661)

diffusion

Title Description

A point every time unit diffuses a distance in the four directions, as shown in FIG.

Tulio

Two points a, b communication, referred to as E (a, b), if and only if the diffusion regions a, b are common. Definition of communication block is arbitrary two points within the block u, v path will exist without fail e (u, a0), e (a0, a1), ..., e (ak, v). Given input format of
the first line of a number n, the n lines, each line point coordinates.

Input Format

[Data] scale

For 20% of the data, to meet 1≤N≤5; 1≤X [i], Y [i] ≤50;

To 100% of the data satisfies 1≤N≤50; 1≤X [i], Y [i] ≤10 ^ 9.

Output Format

A number representing the earliest time point of all the communication block is formed.

N input sample to the point on the output plane, and asked what is the earliest time that they form a communication block.

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Must first know the minimum time into two communicating Harman Dayton two blocks is the distance, which is the most critical;
longest side is required then the n points of minimum spanning tree;

Code:

#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=100100;
const int M=2000100;
const int mod=1e9;
int head[N],cnt,n,x[100],y[100],fa[100];
struct Node{
	int fa,sn,w;
}edge[N*2];
void add(int p,int q,int w){
	edge[cnt].fa=p;
	edge[cnt].sn=q;
	edge[cnt].w=w;
	cnt++;
}
bool cmp(Node p,Node q){return p.w<q.w;}
int find(int p){
	if(p==fa[p]) return p;
	return fa[p]=find(fa[p]);
}
int krus(){
	sort(edge,edge+cnt,cmp);
	int tot=0,ans=0;
	for(int i=0;i<cnt;i++){
		if(find(edge[i].fa)!=find(edge[i].sn)){
			tot++;
			fa[find(edge[i].sn)]=find(edge[i].fa);
			ans=max(edge[i].w,ans);
		}		
		if(tot==n-1) break;
	}
	return ans;
}
int main(){
	memset(head,-1,sizeof(head));
	cin>>n;
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1;i<=n;i++) scanf("%d%d",&x[i],&y[i]);
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			int d=abs(x[j]-x[i])+abs(y[j]-y[i]);
			d=(d+1)/2;
			add(i,j,d),add(j,i,d);
		}
	}
	cout<<krus()<<endl;
	return 0;
}