链接:https://icpcarchive.ecs.baylor.edu/index.php?
option=com_onlinejudge&Itemid=8&page=show_problem&problem=4675
题目大意:
在一个平面上有 n (1<=n<=50) 个矩形。给你左上角和右下角的坐标(0<=x<=10^6, 0<=y<=10^6)。问这些矩形将该平面划分为多少块。
解题思路:
因为n非常小,能够对整个图进行压缩。仅仅要不改变每条边的相对位置。对答案没有影响。
能够将这些矩形的坐标离散化,然后把边上的点标记一下。之后进行简单dfs就可以。(注意离散化的时候,两条边之间至少要隔一个距离)
代码:
/* 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; }
时间: 2024-10-31 13:36:58