Counting Islands II

描述

Country H is going to carry out a huge artificial islands project. The project region is divided into a 1000x1000 grid. The whole project will last for N weeks. Each week one unit area of sea will be filled with land.

As a result, new islands (an island consists of all connected land in 4 -- up, down, left and right -- directions) emerges in this region. Suppose the coordinates of the filled units are (0, 0), (1, 1), (1, 0). Then after the first week there is one island:

#...
....
....
....

After the second week there are two islands:

#...
.#..
....
....

After the three week the two previous islands are connected by the newly filled land and thus merge into one bigger island:

#...
##..
....
....

Your task is track the number of islands after each week's land filling.

输入

The first line contains an integer N denoting the number of weeks. (1 ≤ N ≤ 100000)

Each of the following N lines contains two integer x and y denoting the coordinates of the filled area.  (0 ≤ x, y < 1000)

输出

For each week output the number of islands after that week's land filling.

样例输入
3
0 0
1 1
1 0
样例输出
1
2
1
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e5+;
using namespace std;
using namespace __gnu_cxx;
const int N = ;
int pre[N],ans[][];
int d[][]= {,,,,,-,-,};
int n,cnt=;
set<int>sa;
int Find(int x)
{
if(pre[x] != x)
pre[x] = Find(pre[x]);
return pre[x];
}
void init()
{
for(int i=; i<N; i++)pre[i]=i;
}
int main()
{
init();
int X,Y;
int g=;
cin>>n;
for(int i=; i<n; i++)
{
g++;
cin>>X>>Y;
ans[X][Y]=;
for(int j=; j<; j++)
{
int xx=X+d[j][];
int yy=Y+d[j][];
if(xx>=&&xx<&&yy>=&&yy<)
{
if(ans[xx][yy]==)
{
int aa=Find(xx*+yy),bb=Find(X*+Y);
if(aa!=bb)pre[aa]=bb,g--;
}
}
}
cout<<g<<endl;
}
return ;
}

hihocoder Counting Islands II(并查集)的更多相关文章

  1. Counting Islands II

    Counting Islands II 描述 Country H is going to carry out a huge artificial islands project. The projec ...

  2. Codeforces Round #360 (Div. 1) D. Dividing Kingdom II 并查集求奇偶元环

    D. Dividing Kingdom II   Long time ago, there was a great kingdom and it was being ruled by The Grea ...

  3. [hihoCoder]无间道之并查集

    题目大意: #1066 : 无间道之并查集 时间限制:20000ms 单点时限:1000ms 内存限制:256MB 描述 这天天气晴朗.阳光明媚.鸟语花香,空气中弥漫着春天的气息……额,说远了,总之, ...

  4. hihoCoder 树结构判定(并查集)

    思路:树满足两个条件: 1.顶点数等于边数加一 2.所有的顶点在一个联通块 那么直接dfs或者并查集就可以了. AC代码 #include <stdio.h> #include<st ...

  5. hihocoder 1638:多级并查集

    题目链接 并查集可以用于聚类. import java.io.FileInputStream; import java.io.FileNotFoundException; import java.ut ...

  6. UVA1665 Islands (并查集)

    补题,逆序考虑每个询问的时间,这样每次就变成出现新岛屿,然后用并查集合并统计.fa = -1表示没出现. 以前写过,但是几乎忘了,而且以前写得好丑的,虽然常数比较小,现在重新写练练手.每个单词后面都要 ...

  7. 【占位】HihoCoder 1160 : 攻城略地(并查集好题)

    攻城略地 时间限制:2000ms 单点时限:1000ms 内存限制:256MB 描述 A.B两国间发生战争了,B国要在最短时间内对A国发动攻击.已知A国共有n个城市(城市编号1, 2, …, n),城 ...

  8. HDU 3081 Marriage Match II (网络流,最大流,二分,并查集)

    HDU 3081 Marriage Match II (网络流,最大流,二分,并查集) Description Presumably, you all have known the question ...

  9. HDU 3081 Marriage Match II (二分图,并查集)

    HDU 3081 Marriage Match II (二分图,并查集) Description Presumably, you all have known the question of stab ...

随机推荐

  1. [洛谷P4782]【模板】2-SAT 问题

    题目大意:有$n$个布尔变量 $x_1 \sim x_n$,另有$m$个需要满足的条件,每个条件的形式都是"$x_i$ 为$true/false$或$x_j$为$true/false$&qu ...

  2. “CNKI 中国知网 PDF 全文下载”油猴脚本在线安装地址

    https://greasyfork.org/zh-CN/scripts/18841-cnki-%E4%B8%AD%E5%9B%BD%E7%9F%A5%E7%BD%91-pdf-%E5%85%A8%E ...

  3. 继承spring的validator接口,实现对数据的校验

    在org.springframework.validation这个包中提供了一些对数据校验的方法,其中Validator接口是其中的一个. 现在用Validator接口,完成对数据的校验. 第一步:先 ...

  4. socket编程 ------ 建立 TCP 服务器和客户端流程(阻塞方式)

    服务器端: 服务器端先创建一个socket,然后把这个socket绑定到端口上,接着让它向tcp/ip协议栈请求一个监听服务并创建一个accept队列来接受客户端请求. void creat_tcpS ...

  5. 精通javascript笔记(智能社)——简易tab选项卡及应用面向对象方法实现

    javascript代码(常规方式/面向过程): <script type="text/javascript"> window.onload=function(){ v ...

  6. jsonp应用

    1.服务端jsonp格式数据 如客户想访问 : http://www.runoob.com/try/ajax/jsonp.php?jsonp=callbackFunction. 假设客户期望返回JSO ...

  7. 51nod 拉勾专业算法能力测评消灭兔子 优先队列+贪心

    题目传送门 这道题一开始想了很久...还想着写网络流 发现根本不可能.... 然后就想着线段树维护然后二分什么的 最后发现优先队列就可以了 代码还是很简洁的啦 233 就是把兔子按血量从大到小排序一下 ...

  8. bzoj 2659 几何

    首先考虑(0, 0)到(p, q)这条直线. y = q / p * x. sum{k = 0 to (p - 1) / 2} [q / p * k] 就是直线下方的点数.sum{k = 0 to ( ...

  9. TLS回调函数

    @author: dlive TLS (Thread Local Storage 线程局部存储 )回调函数常用于反调试. TLS回调函数的调用运行要先于EP代码执行,该特性使它可以作为一种反调试技术使 ...

  10. LeetCode 10 Regular Expression Match

    '.' Matches any single character.'*' Matches zero or more of the preceding element. The matching sho ...