3030: ArtWork
时间限制: 4 Sec 内存限制: 128 MB
提交: 43 解决: 17
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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
来源/分类
【题意】
n*m的方格阵，每次可以将一个黑条盖在某一个子矩阵上，保证这个子矩阵窄边为1。每次放上一个黑条后，输出从未被覆盖的格子的连通块数量。
【分析】
考虑并查集维护连通块，为了靠拢并查集的性质，我们把问题逆向思考，从最后的状态开始一个个的揭开黑条。暴力揭即可。
每揭开一个格子，若为空白，暂时先认为出现了一个新连通块，然后考虑去和它四周的空白格子去合并，若合并成功则连通块数量-1
【代码】
#include<bits/stdc++.h>
using namespace std;
const int dir[4][2]={0,1,0,-1,-1,0,1,0};
int mmp[1010][1010],n,m,q,x[10010][2],y[10010][2],ans[10100];
int fa[1010101];
int father(int x){return x==fa[x]?x:fa[x]=father(fa[x]);}
bool join(int x,int y){x=father(x);y=father(y);if(x!=y)return fa[x]=y,true;return false;}
bool vis[1010][1010];
int gid(int x,int y){return (x-1)*m+y;}
void dfs(int x,int y,int dad)
{
if(vis[x][y])return;
vis[x][y]=true;
fa[gid(x,y)]=dad;
for(int i=0;i<4;i++)
{
int dx=x+dir[i][0], dy=y+dir[i][1];
if(dx<1||dy<1||dx>n||dy>m||mmp[dx][dy])continue;
dfs(dx,dy,dad);
}
}
void fuck(int x,int y,int &sig)
{
sig++; //当前是一个空白格子
for(int i=0;i<4;i++)
{
int dx=x+dir[i][0], dy=y+dir[i][1];
if(dx<1||dy<1||dx>n||dy>m||mmp[dx][dy])continue;
if(join(gid(x,y),gid(dx,dy)))sig--; //与邻近的空白融合了
}
}
int main()
{
scanf("%d%d%d",&n,&m,&q);
for(int i=0;i<q;i++)
{
scanf("%d%d%d%d",&x[i][0],&y[i][0],&x[i][1],&y[i][1]);
if(x[i][0]==x[i][1])
{
if(y[i][0]>y[i][1])swap(y[i][0],y[i][1]);
for(int j=y[i][0];j<=y[i][1];j++)
mmp[x[i][0]][j]++;
}
else if(y[i][0]==y[i][1])
{
if(x[i][0]>x[i][1])swap(x[i][0],x[i][1]);
for(int j=x[i][0];j<=x[i][1];j++)
mmp[j][y[i][0]]++;
}
}
for(int i=n*m;i;i--)fa[i]=i;
int sig=0;
memset(vis,false,sizeof(vis));
for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(!mmp[i][j]&&!vis[i][j])
dfs(i,j,gid(i,j)),++sig;
for(int i=q-1;~i;i--)
{
ans[i]=sig;
if(x[i][0]==x[i][1])
{
for(int j=y[i][0];j<=y[i][1];j++)
if(--mmp[x[i][0]][j]==0)fuck(x[i][0],j,sig);
}
else
{
for(int j=x[i][0];j<=x[i][1];j++)
if(--mmp[j][y[i][0]]==0)fuck(j,y[i][0],sig);
}
}
for(int i=0;i<q;i++)printf("%d\n",ans[i]);
}
