ACM学习历程—POJ1151 Atlantis（扫描线 && 线段树）

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 

题目就是求所有矩形的并面积。

通过查阅知道了是扫描线，了解了扫描线的原理，用线段树手写了一下，结果PushUp函数写搓了。。看了AC的代码才知道了原因。

做法就是通过对纵坐标有序化，然后创建区间。

然后通过横向扫描过去，得到每段横向段的高度，乘以宽度就是面积了。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define LL long long

using namespace std;

//线段树
//扫描线
const int maxn = 205;
struct node
{
    int lt, rt;
    double height;
    int num;
}tree[4*maxn];

struct Line
{
    double x;
    double y1, y2;
    bool isLeft;
}line[maxn];

bool cmp(Line a, Line b)
{
    return a.x < b.x;
}

double y[maxn];

//向上更新
void PushUp(int id)
{
    if(tree[id].num > 0)
    {
        tree[id].height = y[tree[id].rt] - y[tree[id].lt];
        return;
    }
    if(tree[id].lt+1 == tree[id].rt)
        tree[id].height = 0;
    else
        tree[id].height = tree[id<<1].height + tree[id<<1|1].height;
}

//建立线段树
void Build(int lt, int rt, int id)
{
    tree[id].lt = lt;
    tree[id].rt = rt;
    tree[id].height = 0;//每段的初值，根据题目要求
    tree[id].num = 0;
    if (lt+1 == rt)
    {
        //tree[id].val = 1;
        return;
    }
    int mid = (lt + rt) >> 1;
    Build(lt, mid, id<<1);
    Build(mid, rt, id<<1|1);
    //PushUp(id);
}

//寻找符合修改的区间通过判断num进行修改
void Updata(int id,Line p)
{
    if(p.y1 <= y[tree[id].lt] && p.y2 >= y[tree[id].rt])
    {
        if (p.isLeft > 0)
            tree[id].num++;
        else
            tree[id].num--;
        PushUp(id);
        return;
    }
    int mid = (tree[id].lt+tree[id].rt) >> 1;
    if (p.y1 < y[mid])
        Updata(id<<1, p);
    if (p.y2 > y[mid])
        Updata(id<<1|1, p);
    PushUp(id);
}

int n;

void Input()
{
    double x1, y1, x2, y2;
    int cnt = 1;
    for (int i = 0; i < n; ++i)
    {
        scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
        y[cnt] = y1;
        y[cnt+1] = y2;

        line[cnt].x = x1;
        line[cnt].y1 = y1;
        line[cnt].y2 = y2;
        line[cnt].isLeft = true;

        line[cnt+1].x = x2;
        line[cnt+1].y1 = y1;
        line[cnt+1].y2 = y2;
        line[cnt+1].isLeft = false;
        cnt += 2;
    }
    sort(y+1, y+1+2*n);
    sort(line+1, line+1+2*n, cmp);
    Build(1, 2*n, 1);
}

double Work()
{
    double ans = 0;
    Updata(1, line[1]);
    int len = 2*n;
    for (int i = 2; i <= len; ++i)
    {
        ans +=  (line[i].x-line[i-1].x) * tree[1].height;
        Updata(1, line[i]);
    }
    return ans;
}

int main()
{
    //freopen("test.in", "r", stdin);
    int times = 1;
    while (scanf("%d", &n) != EOF && n)
    {
        Input();
        double ans = Work();
        printf("Test case #%d
", times);
        printf("Total explored area: %.2lf

", ans);
        times++;
    }
    return 0;
}