矩阵重叠面积计算线段树hdu1542

最新推荐文章于 2022-12-19 14:49:32 发布

转载最新推荐文章于 2022-12-19 14:49:32 发布 · 206 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/mfys/p/7594645.html

文章标签：

#java

本文介绍了一个计算古希腊文献中描述的亚特兰蒂斯不同区域地图总面积的程序设计问题。通过解析输入的地图坐标，利用线段树数据结构来计算所有地图覆盖的总探索面积。

Atlantis

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14378 Accepted Submission(s): 5931

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<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

10 10 20 20

15 15 25 25.5 0

Sample Output

Test case #1

Total explored area: 180.00

转载

线段树的结点是左闭右闭（一般），如果这段区间[L,R] R已经被填补了的话(在r-1地情况下，那么要填补缝隙所以这段就是左闭右+1闭）因为如果单纯地X[r]-X[l]少加了多个节点之间的缝隙，但如果改成X[r+1]-X[l]最后一个又多加了。而查询前先将r-=1两个条件就都能满足了；另外这道题是可以加懒惰标记的。。但是由于离散后范围也很小所以就不加了，。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define lz 2*u,l,mid
#define rz 2*u+1,mid+1,r
const int maxn=4222;
double sum[maxn];
int flag[maxn];
double X[maxn];

struct Node
{
    double lx, rx, y;
    int s;
    Node(){};
    Node(double lx_, double rx_, double y_, int s_)
    {
        lx=lx_, rx=rx_, y=y_, s=s_;
    }
    bool operator <(const Node &S) const
    {
        return y<S.y;
    }
}line[maxn];

int find(double tmp, int n)
{
    int l=1, r=n, mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(X[mid]==tmp) return mid;
        else if(X[mid]<tmp) l=mid+1;
        else r=mid-1;
    }
}

void push_up(int u, int l, int r)
{
    if(flag[u]) sum[u]=X[r+1]-X[l];
    else if(l==r) sum[u]=0;
    else sum[u]=sum[2*u]+sum[2*u+1];
}

void Update(int u, int l, int r, int tl, int tr, int c)
{
    if(tl<=l&&r<=tr)
    {
        flag[u]+=c;
        push_up(u,l,r);
        return ;
    }
    int mid=(l+r)>>1;
    if(tr<=mid) Update(lz,tl,tr,c);
    else if(tl>mid) Update(rz,tl,tr,c);
    else
    {
        Update(lz,tl,mid,c);
        Update(rz,mid+1,tr,c);
    }
    push_up(u,l,r);
}

int main()
{
    int n,tcase=0;
    while(cin >> n,n)
    {
        int num=0;
        memset(flag,0,sizeof(flag));
        memset(sum,0,sizeof(sum));
        for(int i=0; i<n; i++)
        {
            double x1,x2,y1,y2;
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[++num]=Node(x1,x2,y1,1);
            X[num]=x1;
            line[++num]=Node(x1,x2,y2,-1);
            X[num]=x2;
        }
        sort(X+1,X+num+1);
        sort(line+1,line+num+1);
        int k=2*n;
        double ans=0;
        for(int i=1; i<num; i++)
        {
            int l=find(line[i].lx,k);
            int r=find(line[i].rx,k)-1;
            Update(1,1,k,l,r,line[i].s);
            ans+=sum[1]*(line[i+1].y-line[i].y);
        }
        printf("Test case #%d\n",++tcase);
        printf("Total explored area: %.2lf\n\n",ans);
    }
    return 0;
}