3044 矩形面积求并 - Wikioi

题目描述 Description

输入n个矩形，求他们总共占地面积（也就是求一下面积的并）

输入描述 Input Description

可能有多组数据，读到n=0为止(不超过15组)

每组数据第一行一个数n，表示矩形个数(n<=100)

接下来n行每行4个实数x1,y1,x2,y1(0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000)，表示矩形的左下角坐标和右上角坐标

输出描述 Output Description

每组数据输出一行表示答案

样例输入 Sample Input

2
10 10 20 20
15 15 25 25.5
0

样例输出 Sample Output

180.00

水题，我只是拿来练习扫描线的

const
maxn=110;
type
node=record
l,r,lc,rc,cnt:longint;
sum,cover:double;
end;
var
tree:array[0..maxn*4]of node;
x:array[0..maxn*2]of double;
l,r,c:array[0..maxn*2]of longint;
y,lll,rrr:array[0..maxn*2]of double;
n,tot,ll,rr:longint;
ans:double;

procedure swap(var x,y:longint);
var
t:longint;
begin
t:=x;x:=y;y:=t;
end;

procedure swap(var x,y:double);
var
t:double;
begin
t:=x;x:=y;y:=t;
end;

procedure sort(l,r:longint);
var
i,j:longint;
y:double;
begin
i:=l;
j:=r;
y:=x[(l+r)>>1];
repeat
while x[i]<y do
inc(i);
while x[j]>y do
dec(j);
if i<=j then
begin
swap(x[i],x[j]);
inc(i);
dec(j);
end;
until i>j;
if i<r then sort(i,r);
if j>l then sort(l,j);
end;

procedure build(l,r:longint);
var
now,mid:longint;
begin
inc(tot);
now:=tot;
tree[now].l:=l;
tree[now].r:=r;
with tree[now] do
begin
cover:=0;
cnt:=0;
if l=r then
begin
sum:=x[r+1]-x[r];
lc:=0;
rc:=0;
exit;
end;
mid:=(l+r)>>1;
lc:=tot+1;
build(l,mid);
rc:=tot+1;
build(mid+1,r);
sum:=x[r+1]-x[l];
end;
end;

procedure sort2(ll,rr:longint);
var
i,j:longint;
z:double;
begin
i:=ll;
j:=rr;
z:=y[(ll+rr)>>1];
repeat
while y[i]<z do
inc(i);
while y[j]>z do
dec(j);
if i<=j then
begin
swap(l[i],l[j]);
swap(r[i],r[j]);
swap(y[i],y[j]);
swap(c[i],c[j]);
inc(i);
dec(j);
end;
until i>j;
if i<rr then sort2(i,rr);
if j>ll then sort2(ll,j);
end;

function find(k:double):longint;
var
l,r,mid:longint;
begin
l:=1;
r:=n*2;
while l<>r do
begin
mid:=(l+r)>>1;
if x[mid]=k then exit(mid);
if x[mid]>k then r:=mid-1
else l:=mid+1;
end;
exit(l);
end;

procedure init;
var
i:longint;
x1,y1,x2,y2:double;
begin
read(n);
if n=0 then halt;
ans:=0;
tot:=0;
for i:=1 to n do
begin
read(x1,y1,x2,y2);
lll[i*2-1]:=x1;
rrr[i*2-1]:=x2;
y[i*2-1]:=y1;
c[i*2-1]:=1;
lll[i*2]:=x1;
rrr[i*2]:=x2;
y[i*2]:=y2;
c[i*2]:=-1;
x[i*2-1]:=x1;
x[i*2]:=x2;
end;
sort(1,n*2);
build(1,n*2-1);
for i:=1 to n*2 do
begin
l[i]:=find(lll[i]);
r[i]:=find(rrr[i]);
end;
sort2(1,n*2);
end;

procedure insert(now,c:longint);
var
mid:longint;
begin
with tree[now] do
begin
if (ll<=l) and (rr>=r) then
begin
inc(cnt,c);
if cnt>0 then cover:=sum
else cover:=tree[lc].cover+tree[rc].cover;
exit;
end;
mid:=(l+r)>>1;
if ll<=mid then insert(lc,c);
if rr>mid then insert(rc,c);
if cnt>0 then cover:=sum
else cover:=tree[lc].cover+tree[rc].cover;
end;
end;

procedure work;
var
i:longint;
begin
for i:=1 to n*2 do
begin
if y[i]<>y[i-1] then ans:=ans+tree[1].cover*(y[i]-y[i-1]);
ll:=l[i];
rr:=r[i]-1;
insert(1,c[i]);
end;
writeln(ans:0:2);
end;

begin
while true do
begin
init;
work;
end;
end.

View Code