P1661 diffusion-half answer disjoint-set

　　

Title Description

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

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). To n to a point on a given plane, and asked what is the earliest time that they form a communication block.

Input and output formats

Input formats:

 

The first line of a number n, the n lines, each line point coordinates.

[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 formats:

 

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

 

Sample input and output

Input Sample # 1:  Copy

2
0 0
5 5

Output Sample # 1:  Copy

5 

bipartite and answers Manhattan distance <= mid * 2

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int xs[51];
int ys[51];//坐标
int ints[51];//并查集
int find(int n){
    if(ints[n]==n)return(n);
    return(ints[n]=find(ints[n]));
}
int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++)cin>>xs[i]>>ys[i];
    int l=0,r=1000000000;
    int ans=0;//最终答案
    while(l<=r){
        int mid=(l+r)>>1;//二分答案
        for(register int i=0;i<n;i++){
            ints[i]=i;
        }//初始化并查集
        for(register int i=0;i<n;i++){
            for(register intJ = I + . 1 ; J <n-; J ++ ) {
                int DIS = ABS (XS [I] -xs [J]) + ABS (YS [I] - YS [J]);
                IF (DIS <= MID * 2 ) { // can spread to the sides and even 
                   int AA = Find (I), = ab & Find (J);
                    IF (AA = ab &!) INTS [AA] = ab &; 
               } 
            } 
        } 
        int CNT = 0 ; // communication block number 
        for (Register int I = 0 ; I <n-; I ++ ) {
             IF (INTS [I] == I) CNT ++ ; 
        } 
        IF (CNT ==. 1 ) { 
            ANS = MID; 
            R & lt = mid- . 1 ; 
        } // Only one block is updated communication answer look down 
        the else { 
            L = MID + . 1 ; 
        } 
    } 
    COUT << ANS << endl;
     return ( 0 ); 
}

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