POJ 1151 HDU 1542 Atlantis(扫描线)

最新推荐文章于 2020-07-06 20:23:14 发布

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最新推荐文章于 2020-07-06 20:23:14 发布

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原文链接：http://www.cnblogs.com/claireyuancy/p/7160424.html

本文介绍了一种使用扫描线算法解决地图区域叠加问题的方法。通过离散化x坐标并利用线段树进行区间更新，实现了对多个矩形区域重叠部分的总面积计算。

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题目大意就是：去一个地方探险，然后给你一些地图描写叙述这个地方，每一个描写叙述是一个矩形的右下角和左上角。地图有些地方是重叠的。所以让你求出被描写叙述的地方的总面积。

扫描线的第一道题，想了又想，啸爷还给我讲了讲，最终有点理解了啊。

先说扫描线：书上说扫描线不是一个物体。而是一个概念。

在计算几何中的作用类似于图论中的bfs与dfs。所以还是须要多做题目来体会一下啊。

这道题目的做法是：离散化x坐标。然后依照y坐标的大小进行排序，每一条保存它的左边界的位置与右边界的位置。以及自身的高度。

还有就是假设是下边初始为1。上边初始为-1。

接下来就是扫描线了：依照y值排序后的数组，開始遍历。

先二分查找它在离散数组中下表的位置。找到之后按他保存的边界的标记。进行更新。这里的区间更新用的是线段树的维护。我们以离散化后数组建树。

每一个节点保存它此时有多少个上界与下界的和，表示他是否存在矩形。

假设存在的话每一个节点中用sun数组保存这个矩形x值的差值(通过这个离散区间的下标，做减法就是离散的x的差值)。最后的时候乘上y的差值。

就是扫描到的矩阵的面积。

Atlantis

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17207		Accepted: 6549

Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

Input

The input consists of several test cases. Each test case starts with a line containing a single integer n (1 <= n <= 100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don't process it.

Output

For each test case, your program should output one section. The first line of each section must be "Test case #k", where k is the number of the test case (starting with 1). The second one must be "Total explored area: a", where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.

Sample Input

2
10 10 20 20
15 15 25 25.5
0

Sample Output

Test case #1
Total explored area: 180.00

Source

Mid-Central European Regional Contest 2000

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 5010;

struct node
{
    double l, r, h;
    int x;
} f[maxn];

double sum[maxn<<2];
int cnt[maxn];
double dc[maxn];

bool cmp(node a, node b)
{
    return a.h < b.h;
}

int Find(double x, double a[], int n)
{
    int l = 0;
    int r = n-1;
    while(l <= r)
    {
        int mid = (l+r)/2;
        if(a[mid] == x) return mid;
        if(a[mid] > x) r = mid-1;
        else l = mid+1;
    }
    return -1;

}

void Up(int l, int r, int site)
{
    if(cnt[site]) sum[site] = dc[r+1]-dc[l];
    else if(l == r) sum[site] = 0;
    else sum[site] = sum[site<<1]+sum[site<<1|1];
}

void Update(int l, int r, int L, int R, int d, int site)
{
    if(L <= l && r <= R)
    {
        cnt[site] += d;
        Up(l, r, site);
        return;
    }
    int mid = (l+r)>>1;
    if(L <= mid) Update(l, mid, L, R, d, site<<1);
    if(R > mid) Update(mid+1, r, L, R, d, site<<1|1);
    Up(l, r, site);
}

int main()
{
    int n;
    int Case = 1;
    while(cin >>n)
    {
        if(!n) break;
        double x1, y1, x2, y2;
        int m = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf %lf %lf %lf",&x1, &y1, &x2, &y2);
            dc[m] = x1;
            f[m].l = x1;
            f[m].r = x2;
            f[m].h = y1;
            f[m++].x = 1;
            dc[m] = x2;
            f[m].l = x1;
            f[m].r = x2;
            f[m].h = y2;
            f[m++].x = -1;
        }
        sort(dc, dc+m);
        sort(f, f+m, cmp);
        int k = unique(dc, dc+m)-dc;
        memset(cnt, 0 , sizeof(cnt));
        memset(sum, 0 , sizeof(sum));
        double ans = 0;
        for(int i = 0; i < m-1; i++)
        {
            int l = Find(f[i].l, dc, k);
            int r = Find(f[i].r, dc, k)-1;
            if(l <= r) Update(0, k-1, l, r, f[i].x, 1);
            ans += sum[1]*(f[i+1].h-f[i].h);
        }
        printf("Test case #%d\n",Case++);
        printf("Total explored area: %.2f\n\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/claireyuancy/p/7160424.html