POJ 1151 Atlantis 矩阵的并 线段树维护有效值

题意：给定n（n <=100）个矩形的左下角和右上角的坐标（0<=x y<=100000），求覆盖的面积。

题解：把所有的从坐标离散化，横向将所有垂直的线段排序并标记是矩形的左边还是右边线段，扫描时线段树维护有效的长度，累加得到答案。

Sure原创，转载请注明出处。

#include <iostream>
#include <cstdio>
#include <memory.h>
#include <algorithm>
using namespace std;
const int maxn = 102;
struct line
{
double x,down,up;
int left;
bool operator < (const line &other) const
{
return x < other.x;
}
}vertical_line[maxn << 1];
struct node
{
int l,r;
int lazy;
double len;
}seg[maxn << 3];
double yy[maxn << 1];
int n,top,cas = 0;

void read()
{
double x1,x2,y1,y2;
for(int i=0;i<n;i++)
{
scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
vertical_line[i].x = x1;
vertical_line[i].down = y1;
vertical_line[i].up = y2;
vertical_line[i].left = 1;
yy[i] = y1;

vertical_line[i+n].x = x2;
vertical_line[i+n].down = y1;
vertical_line[i+n].up = y2;
vertical_line[i+n].left = -1;
yy[i+n] = y2;
}
sort(vertical_line ,vertical_line + 2 * n);
sort(yy , yy + 2 * n);
top = unique(yy , yy + 2 * n) - yy;
return;
}

void biuld(int l,int r,int num)
{
seg[num].l = l;
seg[num].r = r;
seg[num].lazy = 0;
seg[num].len = 0.0;
if(l + 1 == r) return;
int mid = (l + r) >> 1;
biuld(l , mid , num << 1);
biuld(mid , r , num << 1 | 1);
return;
}

void UP(int num)
{
if(seg[num].lazy > 0)
{
seg[num].len = yy[seg[num].r] - yy[seg[num].l];
}
else if(seg[num].l + 1 == seg[num].r)
{
seg[num].len = 0.0;
}
else
{
seg[num].len = seg[num << 1].len + seg[num << 1 | 1].len;
}
return;
}

void DOWN(int num)
{
if(seg[num].lazy > 0 && seg[num].l + 1 < seg[num].r)
{
seg[num << 1].lazy += seg[num].lazy;
seg[num << 1 | 1].lazy += seg[num].lazy;
seg[num].lazy = 0;
seg[num << 1].len = yy[seg[num << 1].r] - yy[seg[num << 1].l];
seg[num << 1 | 1].len = yy[seg[num << 1 | 1].r] - yy[seg[num << 1 | 1].l];
}
return;
}

void update(int l,int r,int num,int val)
{
if(l == seg[num].l && r == seg[num].r && val + seg[num].lazy >= 0)
{
seg[num].lazy += val;
UP(num);
return;
}
DOWN(num);
int mid = (seg[num].l + seg[num].r) >> 1;
if(mid >= r) update(l , r , num << 1 , val);
else if(l >= mid) update(l , r , num << 1 | 1 , val);
else
{
update(l , mid , num << 1 , val);
update(mid , r , num << 1 | 1 , val);
}
UP(num);
return;
}

void addline(int i)
{
int bjd = lower_bound(yy , yy + top , vertical_line[i].down) - yy;
int bju = lower_bound(yy , yy + top , vertical_line[i].up) - yy;
update(bjd , bju , 1 , vertical_line[i].left);
return;
}

void solve()
{
double ans = 0.0;
addline(0);
for(int i=1;i<2*n;i++)
{
ans += seg[1].len * (vertical_line[i].x - vertical_line[i-1].x);
addline(i);
}
printf("Test case #%d\n",++cas);
printf("Total explored area: %.2f\n\n",ans);
return;
}

int main()
{
while(~scanf("%d",&n) && n)
{
read();
biuld(0,top-1,1);
solve();
}
return 0;
}