UVA 11983 Weird Advertisement --线段树求矩形问题

题意：给出n个矩形，求矩形中被覆盖K次以上的面积的和。

解法：整体与求矩形面积并差不多，不过在更新pushup改变len的时候，要有一层循环，来更新tree[rt].len[i]，其中tree[rt].len[i]表示覆盖次数大于等于i的线段长度，以便求面积，最后只要每次都用tree[1].len[K]来算面积即可。

代码：

#include <iostream>
#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define ll long long
using namespace std;
#define N 30007

struct node
{
    int cov;
    ll len[12];
}tree[8*N];

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

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

void addLine(ll x1,ll x2,ll y1,ll y2,int &m)
{
    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;
}

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

void build(int l,int r,int rt)
{
    tree[rt].cov = 0;
    memset(tree[rt].len,0,sizeof(tree[rt].len));
    if(l+1 == r) 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)
{
    int cov = tree[rt].cov;
    for(int i=0;i<=K;i++)
    {
        if(cov >= i)
            tree[rt].len[i] = yy[r]-yy[l];
        else if(l+1 == r)
            tree[rt].len[i] = 0;
        else
            tree[rt].len[i] = tree[2*rt].len[i-cov] + tree[2*rt+1].len[i-cov];
    }
}

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

int main()
{
    int t,cs = 1,n,m,i,j;
    ll x1,x2,y1,y2;
    yy[0] = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&K);
        m = 1;
        for(i=0;i<n;i++)
        {
            scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2);
            x2++,y2++;
            addLine(x1,x2,y1,y2,m);
        }
        if(K > n)
        {
            printf("Case %d: %d
",cs++,0);
            continue;
        }
        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);
        ll ans = 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 += tree[1].len[K]*(line[i+1].x-line[i].x);
        }
        printf("Case %d: %lld
",cs++,ans);
    }
    return 0;
}

线段树求矩形面积交也可用类似方法，令K = 2即可