题目链接：点击打开链接
题目描述：给定一些矩形，求这些矩形的总面积，如果有重叠，只算一次
解题思路：扫描线+线段树+离散（代码从上往下扫描）
代码：
#include<cstdio>
#include <algorithm>
#define MAXN 110
#define LL ((rt<<1)+1)
#define RR ((rt<<1)+2)
using namespace std;
int n;
struct segment{
double l,r,h;
int f;
bool operator<(const segment& b)const{
return h>b.h;
}
}sg[2*MAXN];
double pos[2*MAXN];
int id;
void addSegment(double x1,double y1,double x2,double y2){
sg[id].l=x1;sg[id].r=x2;
sg[id].h=y1;sg[id].f=1;
pos[id++]=x1;
sg[id].l=x1;sg[id].r=x2;
sg[id].h=y2;sg[id].f=-1;
pos[id++]=x2;
}
int binary(double key,int low,int high){
while(low<=high){
int mid=(low+high)/2;
if(pos[mid]==key)
return mid;
else if(key<pos[mid])
high=mid-1;
else
low=mid+1;
}
return -1;
}
struct Tree{
int l,r;
int cover;
double len;
}tree[8*MAXN];
void build(int rt,int l,int r){
tree[rt].l=l;
tree[rt].r=r;
tree[rt].cover=0;
tree[rt].len=0;
if(l==r-1)
return;
int mid=(l+r)>>1;
build(LL,l,mid);
build(RR,mid,r);
}
void pushup(int rt){
if(tree[rt].cover)
tree[rt].len=pos[tree[rt].r]-pos[tree[rt].l];
else if(tree[rt].l==tree[rt].r-1)
tree[rt].len=0;
else
tree[rt].len=tree[LL].len+tree[RR].len;
}
void update(int rt,int l,int r,int f){
if(tree[rt].l==l&&tree[rt].r==r){
tree[rt].cover+=f;
pushup(rt);
return;
}
int mid=(tree[rt].l+tree[rt].r)>>1;
if(r<=mid)
update(LL,l,r,f);
else if(l>=mid)
update(RR,l,r,f);
else{
update(LL,l,mid,f);
update(RR,mid,r,f);
}
pushup(rt);
}
int main(){
int Case=0;
while(scanf("%d",&n)!=EOF&&n!=0){
id=0;
double x1,y1,x2,y2;
for(int i=0;i<n;++i){
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
addSegment(x1,y1,x2,y2);
}
n=(n<<1);
sort(sg,sg+n);
sort(pos,pos+n);
int m=1;
for(int i=1;i<n;++i)
if(pos[i]!=pos[i-1])
pos[m++]=pos[i];
build(0,0,m-1);
double ans=0;
int l=binary(sg[0].l,0,m-1);
int r=binary(sg[0].r,0,m-1);
update(0,l,r,sg[0].f);
for(int i=1;i<n;i++){
ans+=(sg[i-1].h-sg[i].h)*tree[0].len;
l=binary(sg[i].l,0,m-1);
r=binary(sg[i].r,0,m-1);
update(0,l,r,sg[i].f);
}
printf("Test case #%d\n",++Case);
printf("Total explored area: %.2f\n\n",ans);
}
return 0;
}