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;
}