题意:给出N个矩形,求被这些矩形覆盖K次以上的区域面积。
做过hdu 3642 Get The Treasure和hdu 1255 覆盖的面积这道题应该不难才是。
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <map> #include <algorithm> using namespace std; #define LL(x) (x<<1) #define RR(x) (x<<1|1) #define MID(a,b) (a+((b-a)>>1)) typedef long long LL; const int N=60005; struct Line { int x,y1,y2,flag; Line(){} Line(int a,int b,int c,int d) { x=a;y1=b;y2=c;flag=d; } bool operator<(const Line&b)const { return x<b.x; } }; struct node { int lft,rht; int len[12],flag; int mid(){return MID(lft,rht);} void init(){memset(len,0,sizeof(len));} }; int n,k; vector<int> y; vector<Line> line; map<int,int> H; struct Segtree { node tree[N*4]; void calu(int ind) { if(tree[ind].flag>=k) { int tmp=tree[ind].len[0]; tree[ind].init(); tree[ind].len[k]=tree[ind].len[0]=tmp; } else if(tree[ind].flag>0) { int sum=0,cov=tree[ind].flag; for(int i=1;i<=k;i++) tree[ind].len[i]=0; tree[ind].len[cov]=tree[ind].len[0]; if(tree[ind].lft+1==tree[ind].rht) return; for(int i=1;i<=k;i++) { if(i+cov>=k) tree[ind].len[k]+=tree[LL(ind)].len[i]+tree[RR(ind)].len[i]; else tree[ind].len[i+cov]=tree[LL(ind)].len[i]+tree[RR(ind)].len[i]; } for(int i=cov+1;i<=k;i++) sum+=tree[ind].len[i]; tree[ind].len[cov]-=sum; } else { for(int i=1;i<=k;i++) tree[ind].len[i]=0; if(tree[ind].lft+1==tree[ind].rht) return; for(int i=1;i<=k;i++) tree[ind].len[i]=tree[LL(ind)].len[i]+tree[RR(ind)].len[i]; } } void build(int lft,int rht,int ind) { tree[ind].lft=lft; tree[ind].rht=rht; tree[ind].init(); tree[ind].flag=0; tree[ind].len[0]=y[rht]-y[lft]; if(lft+1!=rht) { int mid=tree[ind].mid(); build(lft,mid,LL(ind)); build(mid,rht,RR(ind)); } } void updata(int st,int ed,int ind,int valu) { int lft=tree[ind].lft,rht=tree[ind].rht; if(st<=lft&&rht<=ed) tree[ind].flag+=valu; else { int mid=tree[ind].mid(); if(st<mid) updata(st,ed,LL(ind),valu); if(ed>mid) updata(st,ed,RR(ind),valu); } calu(ind); } }seg; int main() { int t,t_cnt=0; scanf("%d",&t); while(t--) { y.clear(); H.clear(); line.clear(); scanf("%d%d",&n,&k); for(int i=0;i<n;i++) { int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); x2++;y2++; line.push_back(Line(x1,y1,y2,1)); line.push_back(Line(x2,y1,y2,-1)); y.push_back(y1); y.push_back(y2); } sort(line.begin(),line.end()); sort(y.begin(),y.end()); y.erase(unique(y.begin(),y.end()),y.end()); for(int i=0;i<(int)y.size();i++) H[y[i]]=i; seg.build(0,(int)y.size()-1,1); LL res=0; for(int i=0;i<(int)line.size();i++) { if(i!=0) res+=(LL)(line[i].x-line[i-1].x)*seg.tree[1].len[k]; seg.updata(H[line[i].y1],H[line[i].y2],1,line[i].flag); } printf("Case %d: %lld\n",++t_cnt,res); } return 0; }