[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 305010
int n,q;
int x
;
int y
;
struct SRT{
int v;
int idx;
}z
;
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

,top

;
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]
-left[i][j]);
int tmp=min(top[i-1][j],top
[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
;
bool cmpByX(POINT a,POINT b){
return a.y<b.y||(a.y==b.y && a.x<b.x);
}
int projx
;                           //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
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
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;
}