[Uvalive 6663 count the regions] (DFS + discretization)

Link: https://icpcarchive.ecs.baylor.edu/index.php? Option = com_onlinejudge & Itemid = 8 & page = show_problem & problem = 4675

Question:

There are N (1 <= n <= 50) rectangles on a plane, and the coordinates in the upper left corner and lower right corner (0 <= x <= 10 ^ 6, 0 <= Y <= 10 ^ 6 ). Ask how many blocks the plane is divided by these rectangles.

Solution:

Since N is small, the entire graph can be compressed. As long as the relative position of each edge is not changed, the answer is not affected.

You can discretization the coordinates of these rectangles and mark the points on the edge. Then perform simple DFS. (Note that when discretization, the two sides must be separated by at least one distance)

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;}

[Uvalive 6663 count the regions] (DFS + discretization)