[ACM]CCF CSP [201609-5]E题 祭坛

先离散化，把空间限定在300000*300000的范围内。

做法一：预处理每个点上下左右的节点个数，分分钟写完，时间复杂度O(n^2)，可以拿到【60分】。

做法二：两次线段树。枚举y坐标值，维护当前y坐标上方和下方的节点个数。第一次线段树求出最优结界层数best。如果是询问一，那输出答案。如果是询问二，以best为阙值再构造一个线段树，统计结界层数==best的方案树。代码略长。时间复杂度O(nlogn) 【100分】

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 305010int n,q;int x[N];int y[N];struct SRT{  int v;  int idx;}z[N];bool cmpByV(SRT a,SRT b){  return a.v<b.v;}void reorder(int *x){  for (int i=1;i<=n;i++){    z[i].v=x[i];    z[i].idx=i;  }  sort(z+1,z+n+1,cmpByV);  int newVal=1;  x[z[1].idx]=newVal;  for (int i=2;i<=n;i++){    if (z[i].v!=z[i-1].v)newVal++;    x[z[i].idx]=newVal;  }}int readInt(){  char ch=getchar();  while (ch!='-' &&(ch<'0'||ch>'9'))ch=getchar();  int flag=1;  if (ch=='-'){    flag=-1;    ch=getchar();  }  int x=0;  do{     x=x*10+ch-'0';     ch=getchar();  }while (ch>='0'&&ch<='9');  return x*flag;}void read(){  n=readInt();q=readInt();  for (int i=1;i<=n;i++){    x[i]=readInt();    y[i]=readInt();  }  reorder(x);  reorder(y);  //for (int i=1;i<=n;i++)printf("(%d,%d)\n",x[i],y[i]);  //system("pause");}/*short left[N][N],top[N][N];int min(int a,int b){  return a<b?a:b;}void solve(){  memset(left,0,sizeof(left));  memset(top,0,sizeof(top));  for (int i=1;i<=n;i++) left[x[i]][y[i]]=top[x[i]][y[i]]=1;  for (int i=1;i<=n;i++)    for (int j=1;j<=n;j++){      left[i][j]+=left[i][j-1];    }  for (int j=1;j<=n;j++)  for (int i=1;i<=n;i++)    top[i][j]+=top[i-1][j];  int best=0,best_cnt;  for (int i=1;i<=n;i++)  for (int j=1;j<=n;j++){    int level=min(left[i][j-1],left[i][n]-left[i][j]);    int tmp=min(top[i-1][j],top[n][j]-top[i][j]);    level=min(level,tmp);    if (level>best){      best=level;      best_cnt=1;    }else    if (level==best) best_cnt++;  }  if (q==1)printf("%d\n",best);  else printf("%d\n",best_cnt);}*/int max(int a,int b){return a<b?b:a;}struct POINT{  int x,y;}p[N];bool cmpByX(POINT a,POINT b){  return a.y<b.y||(a.y==b.y && a.x<b.x);}int projx[N];                           //点到x的投影struct NODE{  int l;  int r;  int tval,bval;  //叶子节点信息  int mx;  NODE* ch[2];  NODE(){ch[0]=ch[1]=NULL;}  inline void maintain(){     if (ch[0]==NULL)return ;     mx=max(ch[0]->mx,ch[1]->mx);  }  void build(int ll,int rr){  //按照projx[N]进行初始化     l=ll;     r=rr;     if (l==r-1){       tval=projx[r];       bval=0;       mx=0;       return;     }     ch[0]=new NODE();ch[0]->build(l,(l+r)>>1);     ch[1]=new NODE();ch[1]->build((l+r)>>1,r);     maintain();  }  void insTop(int pos,int v){    if (pos-1>=r ||pos<=l) return;    if (pos-1==l&&pos==r){      tval+=v;      mx=min(tval,bval);      return;    }    ch[0]->insTop(pos,v);    ch[1]->insTop(pos,v);    maintain();  }  void insBot(int pos,int v){    if (pos-1>=r ||pos<=l) return;    if (pos-1==l&&pos==r){      bval+=v;      mx=min(tval,bval);      return;    }    ch[0]->insBot(pos,v);    ch[1]->insBot(pos,v);    maintain();  }  int getMax(int ll,int rr){    if (r<=ll || l>=rr) return 0;    if (l>=ll && r<=rr) return mx;    return max(ch[0]->getMax(ll,rr),ch[1]->getMax(ll,rr));  }  void show(){    if (l==r-1){printf("(%d:%d,%d)",r,tval,bval);return;}    ch[0]->show();ch[1]->show();  }}*root;void pre(){  memset(projx,0,sizeof(projx));  for (int i=1;i<=n;i++){    p[i].x=x[i];    p[i].y=y[i];    projx[x[i]]++;  }  sort(p+1,p+n+1,cmpByX);  //点按y第一x第二排序  root=new NODE();  root->build(0,n);}#define FOR(i,st,ed) for (int i=st;i<=ed;i++)int best;struct NODE2{  int l,r;  NODE2* ch[2];  int tval;  int bval;  int cnt;  //上下均大于best的统计  NODE2(){ch[0]=ch[1]=NULL;}  void maintain(){     if (ch[0]==NULL)return;     cnt=ch[0]->cnt+ch[1]->cnt;  }  void build(int ll,int rr){  //按照projx[N]进行初始化     l=ll;     r=rr;     if (l==r-1){       tval=projx[r];       bval=0;       cnt=0;       return;     }     ch[0]=new NODE2();ch[0]->build(l,(l+r)>>1);     ch[1]=new NODE2();ch[1]->build((l+r)>>1,r);     maintain();  }  void insTop(int pos,int v){    if (pos-1>=r ||pos<=l) return;    if (pos-1==l&&pos==r){      tval+=v;      if (bval>=best&&tval>=best) cnt=1;      return;    }    ch[0]->insTop(pos,v);    ch[1]->insTop(pos,v);    maintain();  }  void insBot(int pos,int v){    if (pos-1>=r ||pos<=l) return;    if (pos-1==l&&pos==r){      bval+=v;      cnt=0;      if (bval>=best&&tval>=best) cnt=1;      return;    }    ch[0]->insBot(pos,v);    ch[1]->insBot(pos,v);    maintain();  }  int count(int ll,int rr){    if (r<=ll || l>=rr) return 0;    if (l>=ll && r<=rr) return cnt;    return ch[0]->count(ll,rr)+ch[1]->count(ll,rr);  }  void show(){    if (l==r-1){printf("(%d:%d %d %d)",r,tval,bval,cnt);return;}    ch[0]->show();ch[1]->show();  }}*root2;void solve2_get_cnt(){  //已知best，求方法数  int curIdx=1;  root2=new NODE2();root2->build(0,n);  int best_cnt=0;  while (curIdx<=n){    int yy=p[curIdx].y;    int stIdx=curIdx;    int edIdx=curIdx;    while (edIdx<=n && p[edIdx+1].y ==yy) edIdx++;    curIdx=edIdx+1;    int pntCnt=edIdx-stIdx+1;    //p[stIdx..edIdx]这些点y坐标均为yy，计算y坐标为yy的ans    for (int i=stIdx;i<=edIdx;i++) root2->insTop(p[i].x,-1);  //从上方减去节点    //printf("root2:");root2->show();putchar('\n');    int p1=stIdx+best-1;    int p2=edIdx-best+1;    if (p1<p2) best_cnt+=root2->count(p[p1].x,p[p2].x-1);    //    for (int i=stIdx;i<=edIdx;i++) root2->insBot(p[i].x,1);  //下方加上    //FOR(i,stIdx,edIdx) printf("(%d,%d) ",p[i].x,p[i].y);putchar('\n');  }  printf("%d\n",best_cnt);}void solve2(){  best=0;  pre();  int curIdx=1;  while (curIdx<=n){    int yy=p[curIdx].y;    int stIdx=curIdx;    int edIdx=curIdx;    while (edIdx<=n && p[edIdx+1].y ==yy) edIdx++;    curIdx=edIdx+1;    int pntCnt=edIdx-stIdx+1;    //p[stIdx..edIdx]这些点y坐标均为yy，计算y坐标为yy的ans    for (int i=stIdx;i<=edIdx;i++) root->insTop(p[i].x,-1);  //从上方减去节点    //printf("root:");root->show();putchar('\n');    int p1=pntCnt/2+stIdx-1;    int p2=edIdx-pntCnt/2+1;    while (p1>=stIdx && p1-stIdx+1>best){      int ans=p1-stIdx+1;      int tbMx=root->getMax(p[p1].x,p[p2].x-1);      //printf("(%d,%d) mx=%d\n",p[p1].x,p[p2].x-1,tbMx);      ans=min(ans,tbMx);      best=max(ans,best);      p1--;      p2++;    }    //    for (int i=stIdx;i<=edIdx;i++) root->insBot(p[i].x,1);  //下方加上    //FOR(i,stIdx,edIdx) printf("(%d,%d) ",p[i].x,p[i].y);putchar('\n');  }  if (q==1){   printf("%d\n",best);   return;  }else  solve2_get_cnt();  //q=2}int main(){  read();  //solve();  solve2();  return 0;}