HDU5654 xiaoxin and his watermelon candy 莫队

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#include<list>

using namespace std;

typedef long long ll;
const int MAXN=200005;
const ll INF=1e9;
const ll mod=1e9+7;

struct nod{
int x1,x2,x3;
void se(int y1,int y2,int y3){
x1=y1;x2=y2;x3=y3;
}
bool const operator<(nod y)const{

if( x1==y.x1 && x2==y.x2  )return x3<y.x3;
else if( x1==y.x1 ) return x2<y.x2;
else return x1<y.x1;
}
};

map<nod,int> mp;

struct Query{
int L,R,id,block;
bool operator < (Query a) const
{ if(block == a.block)
return R<a.R;
return block<a.block;
}
}node[MAXN];

int ans[MAXN];
int m;
int a[MAXN];
int n,unite;
int num[MAXN];
int b[MAXN];

void work(){
int  tmp=0;

int L=1;
int R=0;
int tot=0;int cnt=0;
memset(num,0,sizeof(num));
memset(b,0,sizeof(b));
for(int i=3;i<=n;i++){
if(a[i]>=a[i-1]&&a[i-1]>=a[i-2]){
if( mp[ nod{a[i-2],a[i-1],a[i]} ]==0 ){
mp[ nod{a[i-2],a[i-1],a[i]} ]=++cnt;
}
b[i]=mp[ nod{a[i-2],a[i-1],a[i]} ];
}
}

for(int i=0;i<m;i++){
if( node[i].L>node[i].R ){
ans[ node[i].id ]=0;continue;
}
while( R<node[i].R ){
R++;{
num[ b[R] ]++;
if( num[ b[R] ]==1&&b[R] ){tmp++;}}
}
while( R>node[i].R ){
num[ b[R] ]--;
if( b[R]&&num[b[R]]==0 )tmp--;
R--;
}
while( L<node[i].L ){
num[ b[L] ]--;
if(num[ b[L] ]==0&&b[L])tmp--;
L++;
}
while( L>node[i].L ){
L--;
num[ b[L] ]++;
if( num[ b[L] ]==1&&b[L] )tmp++;
}
ans[ node[i].id ]=tmp;
}

}

int  getint(){
int t = 0 ;
char c = getchar() ;
while(c < '0' || c > '9')  c = getchar() ;
while('0' <= c && c <= '9'){
t = t*10 + c - '0' ;
c = getchar() ;
}
return t ;
}

int main()
{
int t;
scanf("%d",&t);
while(t--){
mp.clear();
scanf("%d",&n);
for(int i=1;i<=n;i++){
a[i]=getint();
}
scanf("%d",&m);
unite=(int ) sqrt(n);
for(int i=0;i<m;i++){
node[i].id=i;
node[i].L=getint();node[i].R=getint();
node[i].L+=2;node[i].R;
node[i].block=node[i].L/unite;
}

sort( node,node+m );
work();
for(int i=0;i<m;i++){
printf("%d\n",ans[i]);
}

}

return 0;
}