[uvalive 6663 count the regions] (DFS + discretization) [easy to understand]

Hello everyone , I meet you again , I'm the king of the whole stack .

The main idea of the topic ：

There are... On a plane n （1<=n<=50) A rectangle . Give you the coordinates of the upper left corner and the lower right corner （0<=x<=10^6, 0<=y<=10^6）. Ask how many blocks these rectangles divide the plane .

Their thinking ：

because n A very small , It can compress the whole graph . Just don't change the relative position of each side . It has no effect on the answer .

The coordinates of these rectangles can be discretized , Then mark the dot on the side . After that, simple dfs Can .（ Pay attention to discretization , There should be at least a distance between the two sides ）

Code ：

/*
ID: [email protected]
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define For(i, n) for (int i = 0; i < n; i++)
using namespace std;
const int MOD = 1000000007;
typedef long long ll;
using namespace std;
struct node {
    int a, b, c, d;
} rec[600];
int x[1200], y[1200];
int xp[1000100], yp[1000100];
int mp[240][240], n;
int lx, ly;
void gao() {
    for (int i = 0; i < n; i++) {
        int A = xp[rec[i].a];
        int B = yp[rec[i].b];
        int C = xp[rec[i].c];
        int D = yp[rec[i].d];
        for (int j = A; j <= C; j++) mp[j][D] = mp[j][B] = 1;
        for (int j = D; j <= B; j++) mp[A][j] = mp[C][j] = 1;
    }
}
int dir[4][2] = {{0, -1}, {0, 1}, {1, 0}, { -1, 0}};
bool in(int x, int y) {
    return (x >= 0 && y >= 0 && x < 2 * lx + 1 && y < 2 * ly + 1);
}
void dfs(int x, int y) {
    mp[x][y] = 1;
    for (int i = 0; i < 4; i++) {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if (in(xx, yy) && !mp[xx][yy]) {
            dfs(xx, yy);
        }
    }
}
int main() {
    while(scanf("%d", &n) != EOF && n) {
        for (int i = 0; i < n; i++) {
            scanf("%d%d%d%d", &rec[i].a, &rec[i].b, &rec[i].c, &rec[i].d);
            x[2 * i] = rec[i].a;
            x[2 * i + 1] = rec[i].c;
            y[2 * i] = rec[i].b;
            y[2 * i + 1] = rec[i].d;
        }
        sort(x, x + 2 * n);
        sort(y, y + 2 * n);
        lx = unique(x, x + 2 * n) - x;
        ly = unique(y, y + 2 * n) - y;
        for (int i = 0; i < lx; i++) {
            xp[x[i]] = 2 * i + 1;
        }
        for (int j = 0; j < ly; j++) {
            yp[y[j]] = 2 * j + 1;
        }
        memset(mp, 0, sizeof(mp));
        gao();
        int fk = 0;
        for (int i = 0; i < 2 * lx; i++) {
            for (int j = 0; j < 2 * ly; j++) if (mp[i][j] == 0) {
                    dfs(i, j);
                    fk++;
                }
        }
        cout << fk << endl;
    }
    return 0;
}

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