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.
4 6 5
2 2 2 6
1 3 4 3
2 5 3 5
4 6 4 6
1 6 4 6
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
#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\n",ans[i]); return 0; }
原文：https://www.cnblogs.com/tetew/p/9749651.html