Atlantis (线段树-离散化-并面积）（基础题）

Atlantis

Problem 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 file 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
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

这题是最基层，但我也看了好久的大佬的博客才看懂：https://blog.csdn.net/qq_18661257/article/details/47622677

题意：计算出所有矩形的面积。

主要思路：把图的上下两条边分别赋值为下边为1，上边为-1，这是为了判断用；

先建个树这里的cover是表示区间是否存在，flag表示上下边1或-1，node表示节点，X[l] ~ X[r]，这里len就表示有效长度了；Edge表示线段y就是与x轴的高，平行与x轴的线；主要看代码：

struct node
{
    int l, r, cover;
    double len;

}N[mn<<2];

struct Edge
{
    double x1, x2, y;
    int flag;
}E[mn<<2];

bool cmp(Edge x, Edge y)
{
    return x.y < y.y;
}

void build(int d, int l, int r)
{
    N[d].l = l, N[d].r = r, N[d].cover = 0,N[d].len = 0.0;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    build(d<<1,l,mid);
    build(d<<1|1,mid+1,r);
}

接下来就是主代码了，缺少点解释以后在补充；

#include 
#include 
#include 
#include 
using namespace std;
#define STOP cout << "STOP!~~" << endl;
const int mn = 1005;
int n, m;
double X[mn];
struct node
{
    int l, r, cover;
    double len;

}N[mn<<2];

struct Edge
{
    double x1, x2, y;
    int flag;
}E[mn<<2];

bool cmp(Edge x, Edge y)
{
    return x.y < y.y;
}

void build(int d, int l, int r)
{
    N[d].l = l, N[d].r = r, N[d].cover = 0,N[d].len = 0.0;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    build(d<<1,l,mid);
    build(d<<1|1,mid+1,r);
}

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

void mmp(int d)
{
    int l = N[d].l - 1, r = N[d].r;//补回原来的
    if (l == r)
        N[d].len = 0;
    else if (N[d].cover)
    {
//        cout << X[r] << " " << X[l] << endl;
        N[d].len = X[r] - X[l];
    }
    else
        N[d].len = N[d<<1].len + N[d<<1|1].len;
}

void update(int d, int ql, int qr, int flag)
{
    int l = N[d].l, r = N[d].r;
    if (ql <= l && qr >= r)
    {
        N[d].cover += flag;
        mmp(d);
        return;
    }
    int mid = (l + r) >> 1;
    if (mid >= ql) update(d<<1,ql,qr,flag);
    if (mid < qr) update(d<<1|1,ql,qr,flag);
    mmp(d);
}

int main()
{
    int i, j, z, cnt = 1;
    double x1, x2, y1, y2;
    while (~scanf("%d",&n)&&n)
    {
        memset(X,0,sizeof(X));
//        memset(N,0,sizeof(N));
        j = 0;
        for (i = 0;i < n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            X[j] = x1;
            E[j].x1 = x1, E[j].x2 = x2, E[j].y = y1;
            E[j++].flag = 1;

            X[j] = x2;
            E[j].x1 = x1, E[j].x2 = x2, E[j].y = y2;
            E[j++].flag = -1;
        }
        sort(X,X+j);
        sort(E,E+j,cmp);
        int k = 1;
        for (i = 1;i < j;i++)
        {
            if (X[i] != X[i-1]) X[k++] = X[i];
        }
        build(1,1,j);//这从1开始
        printf("Test case #%d\n",cnt++);
        double ret = 0.0;
        for (i = 0;i < j;i++)
        {
            int l = Find(j,E[i].x1);
            int r = Find(j,E[i].x2);
            update(1,l+1,r,E[i].flag);//所以这要加1因为是从1开始的
            ret += (E[i+1].y - E[i].y)*N[1].len;
        }
        printf("Total explored area: %.2lf\n\n",ret);
    }
    return 0;
}