Problem A : Counting Squares

From:HDU, 1264

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;
}