HDU 1828 / POJ 1177 Picture --线段树求矩形周长并

题意：给n个矩形，求矩形周长并

解法：跟求矩形面积并差不多，不过线段树节点记录的为：

len: 此区间线段长度

cover: 此区间是否被整个覆盖

lmark,rmark: 此区间左右端点是否被覆盖

num: 此区间分离开的线段的条数

重点在转移的地方，不难理解。

代码：

#include <iostream>
#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 10007

struct node
{
int cov,len,num,lmark,rmark;
}tree[8*N];

struct Line
{
int y1,y2,x;
int cov;
}line[2*N];
int yy[2*N];

int cmp(Line ka,Line kb)
{
return ka.x < kb.x;
}

int bsearch(int l,int r,int x)
{
while(l <= r)
{
int mid = (l+r)/2;
if(x == yy[mid])
return mid;
if(x > yy[mid])
l = mid+1;
else
r = mid-1;
}
return l;
}

void build(int l,int r,int rt)
{
tree[rt].cov = tree[rt].len = tree[rt].num = tree[rt].lmark = tree[rt].rmark = 0;
if(l == r-1) return;
int mid = (l+r)/2;
build(l,mid,2*rt);
build(mid,r,2*rt+1);
}

void pushup(int l,int r,int rt)
{
if(tree[rt].cov)
{
tree[rt].len = yy[r]-yy[l];
tree[rt].lmark = tree[rt].rmark = 1;
tree[rt].num = 1;
}
else if(l+1 == r)
tree[rt].len = tree[rt].num = tree[rt].lmark = tree[rt].rmark = 0;
else
{
tree[rt].len = tree[2*rt].len + tree[2*rt+1].len;
tree[rt].lmark = tree[2*rt].lmark;
tree[rt].rmark = tree[2*rt+1].rmark;
tree[rt].num = tree[2*rt].num+tree[2*rt+1].num-tree[2*rt].rmark*tree[2*rt+1].lmark;
//如果左右区间连续，那么可以合并，减少一个区间
}
}

void update(int l,int r,int aa,int bb,int cover,int rt)
{
if(aa <= l && bb >= r)
{
tree[rt].cov += cover;
pushup(l,r,rt);
return;
}
if(l+1 == r) return;
int mid = (l+r)/2;
if(aa <= mid)
update(l,mid,aa,bb,cover,2*rt);
if(bb > mid)
update(mid,r,aa,bb,cover,2*rt+1);
pushup(l,r,rt);
}

int main()
{
int n,m,cs = 1,i,j;
int x1,y1,x2,y2;
while(scanf("%d",&n)!=EOF)
{
m = 1;
for(i=0;i<n;i++)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
line[m].x = x1,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = 1,yy[m++] = y1;
line[m].x = x2,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = -1,yy[m++] = y2;
}
m--;
sort(yy+1,yy+m+1);
int cnt = 2;
for(i=2;i<=m;i++)
{
if(yy[i] != yy[i-1])
yy[cnt++] = yy[i];
}
cnt--;
build(1,cnt,1);
sort(line+1,line+m+1,cmp);
int ans = 0;
int last = 0;
for(i=1;i<=m;i++)
{
int L = bsearch(1,cnt,line[i].y1);
int R = bsearch(1,cnt,line[i].y2);
update(1,cnt,L,R,line[i].cov,1);
ans += abs(last - tree[1].len);
if(i == m) break;
ans += tree[1].num*2*(line[i+1].x-line[i].x);
last = tree[1].len;
}
printf("%d\n",ans);
}
return 0;
}

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