练习2 - 扩散

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<sstream>
#include <map> 
using namespace std;
#define L 101
#define MAX 0x3f
int m,n;
//int g[L][L];	//占位 
int x,y,k1,k2;
int s,t=0;
int mov[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int wmax,wmin,hmax,hmin;

/*int g(int i,int j){	//模拟负key数组 
	
}*/
int father[L];
int num[L];	//实时统计
int maxl;
bool flag=false;
map<int, map<int, int> > g;	//可以模拟负key数组 => 默认值为0 
map<int, map<int, int> > gg;	//本轮感染不再染 

int find(int k){
	if(father[k]!=k)father[k] = find(father[k]);
	return father[k];
} 
void unity(int x,int y){
	k1=find(x);
	k2=find(y);
	if(k1!=k2){
		father[k1]=k2;	//合并代表
		num[k1]+=num[k2];
		maxl = max(maxl,num[k1]);
		if(maxl>=n) flag = true;
	}
	
} 

int main()
{	
	cin >> n; 
	for(int i=1;i<=n;i++){
		cin >> x >> y;
		g[x][y]=i;	//初始占位 
		wmax=max(wmax,x);	//记录最大边界 
		hmax=max(hmax,y);
		wmin=min(wmin,x);
		hmin=min(hmin,y);
		father[i]=i;
		num[i]=1;	
	}
	while(1){
		t++;	//累加时间 
		for(int i=wmin;i<=wmax;i++){
			for(int j=hmin;j<=hmax;j++){
				if(g[i][j] && !gg[i][j]){	//已经占领了 => 扩张 
					for(int k=0;k<=3;k++){
						x = i+mov[k][0];
						y = j+mov[k][1];
						if(!g[x][y]){g[x][y] = g[i][j];gg[x][y]=1;}	//扩张
						else{unity(g[i][j],g[x][y]);}
					} 
				}
				if(flag){cout << t;return 0;}
			}
		} 
		gg.erase(gg.begin(),gg.end());
		wmax+=1;	//必定扩展边界
		hmax+=1; 
		wmin-=1;
		hmin-=1;
	} 
	
	return 0; 
}
/*
2
0 0
5 5
*/
//cout << "输入："

结果：