问题 A: ArtWork
时间限制: 4 Sec 内存限制: 128 MB
提交: 18 解决: 6
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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.
输入
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 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
#include<bits/stdc++.h>
using namespace std;
struct stock{
int x1,x2,y1,y2;
}pos[10010];
const int N=1010;
int num[N*N],fa[N*N],ans[10010];
int dir[4][2]={0,1,-1,0,0,-1,1,0};
int m,n,p,t;
int hsh(int x,int y){
int num=(x-1)*m+y;
return num;
}
void init(){
for(int i=1;i<=n*m;i++){
fa[i]=i;
num[i]=0;
}
}
int find(int x){
return x==fa[x]?x:fa[x]=find(fa[x]);
}
void merge(int x,int y){
int fx=find(x),fy=find(y);
if(fx==fy)return;
t--;
fa[fx]=fy;
}
bool check(int x,int y){
if(x>=1&&y>=1&&x<=n&&y<=m)
return true;
return false;
}
void work(int x,int y){
for(int i=0;i<4;i++){
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(check(xx,yy)&&!num[hsh(xx,yy)]){
merge(hsh(xx,yy),hsh(x,y));
}
}
}
int main() {
scanf("%d%d%d",&n,&m,&p);
t=n*m;
init();
for(int i=1;i<=p;i++){
scanf("%d%d%d%d",&pos[i].x1,&pos[i].y1,&pos[i].x2,&pos[i].y2);
for(int x=pos[i].x1;x<=pos[i].x2;x++)
for(int y=pos[i].y1;y<=pos[i].y2;y++){
if(num[hsh(x,y)]==0)t--;
num[hsh(x,y)]++;
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if(!num[hsh(i,j)])
work(i,j);
for(int i=p;i>0;i--){
ans[i]=t;
for(int x=pos[i].x1;x<=pos[i].x2;x++)
for(int y=pos[i].y1;y<=pos[i].y2;y++){
num[hsh(x,y)]--;
if(num[hsh(x,y)]==0){
t++;
work(x,y);
}
}
}
for(int i=1;i<=p;i++){
printf("%d\n",ans[i]);
}
return 0;
}