hdu1542 Atlantis--扫描线

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;
    else
      r = 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);
  else
    update(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;
}