Description
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(x2,y2)(x1=x2 or y1=y2)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,x2x1,y1,x2 and y2(1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ m). Either x1 = x2 or y1=y2y1=y2(or both).
Output
For each of the q strokes, output a line containing the beauty of the artwork after the stroke.
Sample Input
4 6 5 2 2 2 6 1 3 4 3 2 5 3 5 4 6 4 6 1 6 4 6
Sample Output
1 3 3 4 3
Hint
Source
NCPC 2016
#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;
}