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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 longusing namespace std;#define N 30007struct 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;}
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线段树求矩形面积交也可用类似方法,令K = 2即可

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