csuoj2022-Artwork（并查集连通块）

Description

点击打开题目链接

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 (x₁, y₁), ends at square (x2,y2)(x1=x2 or y1=y2) and changes the color of all squares (x, y) to black where x₁ ≤ x ≤ x₂ and y₁ ≤ y ≤ y₂. 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.

Input

The first 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 integersx1,y1,x2 and y₂(1 ≤ x₁ ≤ x₂ ≤ n, 1 ≤ y₁ ≤ y₂ ≤ m). Either x₁ = x₂ or y1=y2(or both).

Output

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

Sample Input

Sample Output

Hint

Source

NCPC 2016

题目大意：给m*n的网格，q次操作和询问，每次将两个格子间的全部格子涂黑（只有竖直和水平两种情况），然后询问剩下的白色连通块的个数

思路分析：离线记录每次操作，从末状态倒跑回初始状态，，用并查集记录白块的连通情况，每次判断连通情况和修改连通块的个数即可

代码

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int A[1010][1010],n,m,p[1000010],B[10010],C[10010],D[10010],E[10010],dx[]={0,0,-1,1},dy[]={-1,1,0,0},cnt;
                                  //A记录每点涂黑的次数，B,C,D,E记录每次操作时两点的坐标位置，cnt记录连通块数量
int find(int x){
   if(x==p[x]) return x;
   int k=find(p[x]);
   p[x]=k;
   return k;
}
void dfs(int x,int y,int k){   //末状态时统计连通块个数
     p[m*(x-1)+y]=k;
     int u,v,i;
     for(i=0;i<=3;i++){
          u=x+dx[i];
          v=y+dy[i];
          if(u>=1&u<=n&&v>=1&&v<=m&&A[u][v]==0&&p[m*(u-1)+v]==0) dfs(u,v,k);
     }
}
void unio(int x,int y){
    int i,u,v,f1,f2,flag=0;
    p[m*(x-1)+y]=m*(x-1)+y;
    cnt++;                //合并操作，对于该点判断是否和周围四点连通，可以合并时联通块数量-1；
    u=x+dx[i];
    for(i=0;i<=3;i++){
        v=y+dy[i];
        if(u>=1&&v>=1&&u<=n&&v<=m&&!A[u][v]){
                f1=find(m*(x-1)+y);
                f2=find(m*(u-1)+v);
                if(f1!=f2){
                    cnt--;
                    p[f2]=f1;
                }
        }
    }
}
int main(){
    int i,j,k,q,x1,y1,x2,y2,ans[10010];
    scanf("%d%d%d",&m,&n,&q);
    memset(A,0,sizeof(A));
    memset(p,0,sizeof(p));
    for(i=1;i<=q;i++){
            scanf("%d%d%d%d",&y1,&x1,&y2,&x2);
            B[i]=x1;
            C[i]=y1;
            D[i]=x2;
            E[i]=y2;
            if(x1==x2){
                if(y1>y2) for(j=y2;j<=y1;j++) A[x1][j]++;
                else for(j=y1;j<=y2;j++) A[x1][j]++;
            }
            else{
                if(x1>x2) for(j=x2;j<=x1;j++) A[j][y1]++;
                else for(j=x1;j<=x2;j++) A[j][y1]++;
            }
    }
    cnt=0;
    for(i=1;i<=n;i++){
        for(j=1;j<=m;j++){
            if(A[i][j]==0&&p[m*(i-1)+j]==0){
                    cnt++;
                    k=m*(i-1)+j;
                    dfs(i,j,k);
            }
        }
    }

    //倒跑合并过程
    ans[q]=cnt;
    for(i=q;i>=1;i--){
        if(B[i]==D[i]){
            if(C[i]>E[i]){
                for(j=E[i];j<=C[i];j++){
                     A[B[i]][j]--;
                     if(A[B[i]][j]==0) unio(B[i],j);
                }
            }
            else{
                for(j=C[i];j<=E[i];j++){
                     A[B[i]][j]--;
                     if(A[B[i]][j]==0) unio(B[i],j);
                }
            }
        }
        else{
             if(B[i]>D[i]){
                for(j=D[i];j<=B[i];j++){
                    A[j][C[i]]--;
                    if(A[j][C[i]]==0) unio(j,C[i]);
                }
             }
             else{
                 for(j=B[i];j<=D[i];j++){
                    A[j][C[i]]--;
                    if(A[j][C[i]]==0) unio(j,C[i]);
                 }
             }
        }
        ans[i-1]=cnt;
    }
    for(i=1;i<=q;i++) cout<<ans[i]<<endl;
    return 0;
}