Artwork Gym - 101550A

Posted 2022-09-28 stay-online

题目

　　https://cn.vjudge.net/contest/323505#problem/A

题意

　　给出一个m×n的矩阵，一开始全是白色，然后给出q次操作，每次将（x1，y1）到（x2，y2）这宽度为1的区域涂黑。输出每次涂黑操作后白色可以分成多少块。

题解

　　一开始从各种方向去想BFS的优化QAQ，后来才知道是一道并查集的题目。我们可以反向思考，先将每次涂黑的操作进行完毕，每次给涂黑的地方+1，然后把最后状态的白色块用并查集连到一起。那么我们反向删除黑色部分，当删除的这一块权值变为0时，意味着这一块从黑变成白了，那么我们先将其当作独立的一块，记录num++，然后去跑这一点的上下左右，如果两点分属不同的集合，那么num--，并且将外面的点以当前点为父亲连入。每次操作后记录下来答案，最后输出即可。学到了学到了qwq

 

 #include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#define ull unsigned long long
#define met(a, b) memset(a, b, sizeof(a))
#define lowbit(x) (x&(-x))
#define MID (l + r) / 2
#define ll long long

using namespace std;

const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e6 + 3;
const int maxn = 1e4 + 7;
const int N = 1010;

struct Que 
    int x1, x2, y1, y2;
que[maxn];

int MAP[N][N];
int dx[] = 0, 0, -1, 1;
int dy[] = -1, 1, 0, 0;
int f[1000100];
int vis[1000100];
int res[1000100];

int Find(int i) return f[i] = (f[i] == i) ? f[i] : f[i] = Find(f[i]);

int main() 
    int n, m, q;
    cin >> m >> n >> q;
    for(int i = 1; i <= n*m+2; i++) f[i] = i;
    for(int i = 1; i <= q; i++) 
        cin >> que[i].y1 >> que[i].x1 >> que[i].y2 >> que[i].x2;
        if(que[i].x1 == que[i].x2) 
            int t1 = min(que[i].y1, que[i].y2);
            int t2 = max(que[i].y1, que[i].y2); 
            for(int j = t1; j <= t2; j++) MAP[que[i].x1][j]++;
        
        else 
            int t1 = min(que[i].x1, que[i].x2);
            int t2 = max(que[i].x1, que[i].x2);
            for(int j = t1; j <= t2; j++) MAP[j][que[i].y1]++;
        
    
    // for(int i = 1; i <= n; i++) 
    //     for(int j = 1; j <= m; j++) 
    //         cout << MAP[i][j] << ‘ ‘;
    //     
    //     cout << endl;
    // 
    for(int i = 1; i <= n; i++) 
        for(int j = 1; j <= m; j++) 
            if(!MAP[i][j]) 
                int id = (i-1)*m+j;
                for(int k = 0; k < 4; k++) 
                    int x = i+dx[k];
                    int y = j+dy[k];
                    if(x < 1 || x > n) continue;
                    if(y < 1 || y > m) continue;
                    if(!MAP[x][y]) 
                        int t = (x-1)*m+y;
                        int fi = Find(id);
                        int fo = Find(t);
                        f[fo] = fi;
                    
                
            
        
    
    int num = 0;
    for(int i = 1; i <= n; i++) 
        for(int j = 1; j <= m; j++) 
            int fi = Find((i-1)*m+j);
            if(!MAP[i][j] && !vis[fi]) 
                vis[fi] = 1;
                num++;
            
        
    
    for(int i = q; i >= 1; i--) 
        res[i] = num;
        if(que[i].x1 == que[i].x2) 
            int t1 = min(que[i].y1, que[i].y2);
            int t2 = max(que[i].y1, que[i].y2); 
            for(int j = t1; j <= t2; j++) 
                MAP[que[i].x1][j]--;
                if(!MAP[que[i].x1][j]) 
                    num++;
                    int id = (que[i].x1-1)*m+j;
                    for(int k = 0; k < 4; k++) 
                        int x = que[i].x1+dx[k];
                        int y = j+dy[k];
                        if(x < 1 || x > n) continue;
                        if(y < 1 || y > m) continue;
                        if(!MAP[x][y]) 
                            int t = (x-1)*m+y;
                            int fi = Find(id);
                            int fo = Find(t);
                            if(fi != fo) num--;
                            f[fo] = fi;
                        
                    
                
            
        
        else 
            int t1 = min(que[i].x1, que[i].x2);
            int t2 = max(que[i].x1, que[i].x2);
            for(int j = t1; j <= t2; j++) 
                MAP[j][que[i].y1]--;
                if(!MAP[j][que[i].y1]) 
                    num++;
                    int id = (j-1)*m+que[i].y1;
                    for(int k = 0; k < 4; k++) 
                        int x = j+dx[k];
                        int y = que[i].y1+dy[k];
                        if(x < 1 || x > n) continue;
                        if(y < 1 || y > m) continue;
                        if(!MAP[x][y]) 
                            int t = (x-1)*m+y;
                            int fi = Find(id);
                            int fo = Find(t);
                            if(fi != fo) num--;
                            f[fo] = fi;
                        
                    
                
            
        
    
    for(int i = 1; i <= q; i++) cout << res[i] << endl;
    return 0;