NCPC2016-A-ArtWork

Posted tetew

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了NCPC2016-A-ArtWork相关的知识,希望对你有一定的参考价值。

题目描述

A template for an artwork is a white grid of n × m squares. The artwork will be created by painting q horizontal and vertical black strokes. A stroke starts from square (x 1 , y 1 ), ends at square (x 2 , y 2 ) (x 1 = x 2 or y 1 = y 2 ) and changes the color of all squares (x, y) to black where
x 1 ≤ x ≤ x 2 and y 1 ≤ y ≤ y 2 .

The beauty of an artwork is the number of regions in the grid. Each region consists of one or more white squares that are connected to each other using a path of white squares in the grid, walking horizontally or vertically but not diagonally. The initial beauty of the artwork is 1. Your task is to calculate the beauty after each new stroke. Figure A.1 illustrates how the beauty of the artwork varies in Sample Input 1.

技术分享图片

输入

The ?rst line of input contains three integers n, m and q (1 ≤ n, m ≤ 1000, 1 ≤ q ≤ 104 ).
Then follow q lines that describe the strokes. Each line consists of four integers x 1 , y 1 , x 2 and y 2 (1 ≤ x 1 ≤ x 2 ≤ n, 1 ≤ y 1 ≤ y 2 ≤ m). Either x 1 = x 2 or y 1 = y 2 (or both).

输出

For each of the q strokes, output a line containing the beauty of the artwork after the stroke.

样例输入

4 6 5
2 2 2 6
1 3 4 3
2 5 3 5
4 6 4 6
1 6 4 6

样例输出

1
3
3
4
3
题意是给你一个n*m的矩阵,q次询问,每次将连续的一些竖直或水平的格子染黑,问每一步操作之后白色联通块的个数

从最后一种局面往前走,先求出所有操作之后白色联通块的数量,然后逐条删去黑线,对新出现的白格子,要么和原有的某个联通块相连,要么属于单独的联通块
用并查集维护联通块
技术分享图片
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e3+10;
const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};
int n,m,q,cnt;
struct line{
    int x1,x2,y1,y2;}li[N*10];
int f[N*N],num[N][N],ans[N*10];
int Hash(int x,int y)
{
    return (x-1)*m+y;
}
int fund(int x)
{
    if (f[x]==x) return f[x];
    return f[x]=fund(f[x]);
}
void join(int x,int y)
{
    int fx=fund(x),fy=fund(y);
    if (fx!=fy)
    {
        cnt--;
        f[fx]=fy;
    }
}
void dfs(int x,int y)
{
    int id=Hash(x,y);
    for (int i=0;i<4;i++)
    {
        int fx=x+dx[i],fy=y+dy[i];
        if (fx<1||fx>n||fy<1||fy>m) continue;
        if (num[fx][fy]==0)
        {
            join(id,Hash(fx,fy));
        }
    }
}
void print(line l)
{
    for (int i=l.x1;i<=l.x2;i++)
    for (int j=l.y1;j<=l.y2;j++)
    {
        if (num[i][j]==0) cnt--;
        num[i][j]++;
    }
}
void reprint(line l)
{
    for (int i=l.x1;i<=l.x2;i++)
    for (int j=l.y1;j<=l.y2;j++)
    {
        num[i][j]--;
        if (num[i][j]==0)
        {
            cnt++;
            dfs(i,j);
        }
    }
}

int main()
{
    scanf("%d%d%d",&n,&m,&q);
    cnt=n*m;
    for (int i=1;i<=cnt;i++) f[i]=i;
    for (int i=1;i<=q;i++)
    {
        scanf("%d%d%d%d",&li[i].x1,&li[i].y1,&li[i].x2,&li[i].y2);
        print(li[i]);
    }

    for (int i=1;i<=n;i++)
    for (int j=1;j<=m;j++)
        if (num[i][j]==0)  dfs(i,j);

    for (int i=q;i>=1;i--)
    {
        ans[i]=cnt;
        reprint(li[i]);
    }
    for (int i=1;i<=q;i++) printf("%d
",ans[i]);
    return 0;
}
View Code

 

 

以上是关于NCPC2016-A-ArtWork的主要内容,如果未能解决你的问题,请参考以下文章

NCPC 2015 October 10, 2015 Problem D

java 2017 NCPC问题G Chaining

NCPC 2016:简单题解

2019.09.29NCPC2018

NCPC2016-E-Exponial

Fleecing the Raffle(NCPC 2016 暴力求解)