HDU 1264 Counting Squares （线段树-扫描线-矩形面积并）

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Problem A : Counting Squares

pid=1264" rel="nofollow">From:HDU, 1264

Problem Description

Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line
5 8 7 10
specifies the rectangle who's corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.

Input

The input format is a series of lines, each containing 4 integers. Four -1's are used to separate problems, and four -2's are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of
a rectangle.

Output

Your output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line.

Sample Input

5 8 7 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2

Sample Output

8
10000

Source

浙江工业大学第四届大学生程序设计竞赛

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题目大意：

 给定你一些矩形左下右上角坐标点。或者左上右下坐标点。求这些矩形的面积并。

解题思路：

利用线段树扫描线的知识。此题不须要离散化。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

struct node{
    int x,y1,y2,c;
    node(int x0=0,int y10=0,int y20=0,int c0=0){
        x=x0;y1=y10;y2=y20;c=c0;
    }
    friend bool operator < (node a,node b){
        if(a.x!=b.x) return a.x<b.x;
        else if(a.y1!=b.y1) return a.y1<b.y1;
        else if(a.y2!=b.y2) return a.y2<b.y2;
        else return a.c>b.c;
    }
};

const int maxh=110;

struct tree{
    int l,r,c,lz;
}a[maxh*4];

vector <node> v;

bool input(){
    int a,b,c,d;
    v.clear();
    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){
        if(a==-1 && b==-1 && c==-1 && d==-1) return true;
        if(a==-2 && b==-2 && c==-2 && d==-2) return false;
        v.push_back(node( min(a,c), min(b,d) , max(b,d) ,1));
        v.push_back(node( max(a,c), min(b,d) , max(b,d) ,-1));
    }
}

void build(int l,int r,int k){
    a[k].l=l;
    a[k].r=r;
    a[k].c=0;
    a[k].lz=0;
    if(l+1<r){
        int mid=(l+r)/2;
        build(l,mid,2*k);
        build(mid,r,2*k+1);
    }
}

void pushdown(int k){
    if(a[k].lz!=0 && a[k].l+1<a[k].r ){
        a[2*k].lz+=a[k].lz;
        a[2*k+1].lz+=a[k].lz;
        a[2*k].c+=a[k].lz;
        a[2*k+1].c+=a[k].lz;
        a[k].lz=0;
    }
}

void insert(int l,int r,int k,int c){
    if(l<=a[k].l && a[k].r<=r){
        a[k].lz+=c;
        a[k].c+=c;
    }else{
        pushdown(k);
        int mid=(a[k].l+a[k].r)/2;
        if(r<=mid) insert(l,r,2*k,c);
        else if(l>=mid) insert(l,r,2*k+1,c);
        else{
            insert(l,mid,2*k,c);
            insert(mid,r,2*k+1,c);
        }
    }
}

int query(int l,int r,int k){
    pushdown(k);
    if(l<=a[k].l  && a[k].r<=r){
        if(a[k].c>0) return r-l;
        else{
            if(a[k].l+1==a[k].r) return 0;
            else {
                int mid=(a[k].l+a[k].r)/2;
                return query(l,mid,2*k) + query(mid,r,2*k+1) ;
            }
        }
    }else{
        int mid=(a[k].l+a[k].r)/2;
        if(r<=mid) return query(l,r,2*k);
        else if(l>=mid) return query(l,r,2*k+1);
        else{
            return query(l,mid,2*k) + query(mid,r,2*k+1) ;
        }
    }
}

void solve(){
    build(0,maxh,1);
    sort(v.begin(),v.end());
    insert(v[0].y1,v[0].y2,1,v[0].c);
    int ans=0;
    for(int i=1;i<v.size();i++){
        //cout<<v[i].x-v[i-1].x<<" "<<query(0,maxh,1)<<endl;
        ans+=(v[i].x-v[i-1].x)*query(0,maxh,1);
        insert(v[i].y1,v[i].y2,1,v[i].c);
    }
    cout<<ans<<endl;
}

int main(){
    while(input()){
        solve();
    }
    solve();
    return 0;
}