[UVALive 6663 Count the Regions] (dfs + 离散化）

来源：互联网发布：量化交易程序员招聘编辑：程序博客网时间：2024/06/05 04:12

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：

由于n很小，可以对整个图进行压缩，只要不改变每条边的相对位置，对答案没有影响。

可以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即可。（注意离散化的时候，两条边之间至少要隔一个距离）

代码：

/*ID: wuqi9395@126.comPROG: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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