HDU 5172 GTY & amp; #39; s gay friends (line segment tree)

HDU 5172 GTY & #39; s gay friends (line segment tree)

Address: HDU 5172

During the competition, the water of the Line Segment tree that maintains the Range Sum, the range product, and the maximum value of the range passed .. After the game, the data will be changed back to 10 ^ 6, and then it will be TLE ..

The positive solution is the interval and the prefix and maintenance. Then, maintain a occurrence location on the number. If the previous occurrence location of each number in each query is smaller than l, it indicates that there are no duplicates, if the range and conform to the sum in the full arrangement, it indicates that it must be a full arrangement.

The Code is as follows:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include #include 
     
      #include 
      #include 
       
        #include 
        
         using namespace std;#define LL __int64#define pi acos(-1.0)const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;#define root 1, n, 1#define lson l, mid, rt<<1#define rson mid+1, r, rt<<1|1int a[1000002], pre[1000002], pos[1000002];int Max[4000000];LL sum[1000002];void PushUp(int rt){ Max[rt]=max(Max[rt<<1],Max[rt<<1|1]);}void build(int l, int r, int rt){ if(l==r){ Max[rt]=pre[l]; return ; } int mid=l+r>>1; build(lson); build(rson); PushUp(rt);}int Query(int ll, int rr, int l, int r, int rt){ if(ll<=l&&rr>=r){ return Max[rt]; } int mid=l+r>>1, ans=-1; if(ll<=mid) ans=max(ans,Query(ll,rr,lson)); if(rr>mid) ans=max(ans,Query(ll,rr,rson)); return ans;}int main(){ int n, m, i, j, x, l, r, ans; while(scanf("%d%d",&n,&m)!=EOF){ memset(pos,-1,sizeof(pos)); sum[0]=0; for(i=1;i<=n;i++){ scanf("%d",&x); pre[i]=pos[x]; pos[x]=i; sum[i]=sum[i-1]+x; } build(root); while(m--){ scanf("%d%d",&l,&r); if(sum[r]-sum[l-1]!=(LL)(r-l+1)*(r-l+2)/2){ puts("NO"); continue ; } ans=Query(l,r,root); if(ans