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[*N];

struct Line
{
    int y1,y2,x;
    int cov;
}line[*N];
int yy[*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)/;
        if(x == yy[mid])
            return mid;
        if(x > yy[mid])
            l = mid+;
        else
            r = mid-;
    }
    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 = ;
    if(l == r-) return;
    int mid = (l+r)/;
    build(l,mid,*rt);
    build(mid,r,*rt+);
}

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 = ;
        tree[rt].num = ;
    }
    else if(l+ == r)
        tree[rt].len = tree[rt].num = tree[rt].lmark = tree[rt].rmark = ;
    else
    {
        tree[rt].len = tree[*rt].len + tree[*rt+].len;
        tree[rt].lmark = tree[*rt].lmark;
        tree[rt].rmark = tree[*rt+].rmark;
        tree[rt].num = tree[*rt].num+tree[*rt+].num-tree[*rt].rmark*tree[*rt+].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+ == r) return;
    int mid = (l+r)/;
    if(aa <= mid)
        update(l,mid,aa,bb,cover,*rt);
    if(bb > mid)
        update(mid,r,aa,bb,cover,*rt+);
    pushup(l,r,rt);
}

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