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\n",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\n",cs++,ans);

    }

    return 0;

}

View Code

 

线段树求矩形面积交也可用类似方法，令K = 2即可