hdu1542 Atlantis--扫描线

来源：互联网发布：天津网络作家排行榜编辑：程序博客网时间：2024/05/17 14:30

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=1542

题意：

给定n个矩形，n行数据，每行四个，代表矩形的左下角，右上角坐标，问矩形的总面积。就是求矩形并面积。

思路：

以横坐标个数为区间建线段树，定义一个结构体做扫描线。

#define _CRT_SECURE_NO_DEPRECATE #include<iostream>#include<algorithm>#include<cmath>#include<string.h>using namespace std;struct Line{double l, r, h;int flag;bool operator<(Line & a) const{return h < a.h;}}line[405 * 4];int n;int cover[405 * 4];//覆盖的次数double X[405 * 4];//保存所有横坐标double sum[405];//底边长/* 二分搜索 */int BS(double x){int l = -1;int r = n;int mid;while (1){mid = (l + r) / 2;if (X[mid] == x)return mid;else if (X[mid] < x)l = mid;elser = mid;if (r - l == 1)return r;}return -1;}void update(int pos, int flag, int l, int r, int root){if (l == r){cover[pos] += flag;if (cover[pos])sum[root] = X[pos + 1] - X[pos];if (!cover[pos])sum[root] = 0;return;}int mid = (l + r) / 2;if (pos <= mid)update(pos, flag, l, mid, root * 2);elseupdate(pos, flag, mid + 1, r, root * 2 + 1);sum[root] = sum[root * 2] + sum[root * 2 + 1];}int main(){int num;int T = 1;double ans;double x1, y1, x2, y2;while (~scanf("%d", &n) && n){num = 1;int t = 1;for (int i = 0; i < n; i++){scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);line[t].h = y1;line[t].l = x1;line[t].r = x2;line[t++].flag = 1;line[t].h = y2;line[t].l = x1;line[t].r = x2;line[t++].flag = -1;X[num++] = x1;X[num++] = x2;}n = t;sort(X + 1, X + num);sort(line + 1, line + n);memset(sum, 0, sizeof(sum));memset(cover, 0, sizeof(cover));ans = 0.0;for (int i = 1; i <= n; i++){ans += sum[1] * (line[i].h - line[i - 1].h);int l = BS(line[i].l);int r = BS(line[i].r) - 1;for (int j = l; j <= r; j++)update(j, line[i].flag, 1, n, 1);}printf("Test case #%d\nTotal explored area: %.2f\n\n", T++, ans);}return 0;}

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