BZOJ2090 : [Poi2010]Monotonicity 2

设f[i]表示以i为结尾的最长的合法序列的长度，=号直接维护，<号和>号用两棵树状数组维护即可，时间复杂度$O(n\log n)$。

 

#include<cstdio>
#define N 1000000
int n,k,i,j,a[N],e[N+1],bl[N+1],bg[N+1],f[N],ans;char s[N];
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
inline void read(char&a){while(((a=getchar())!='>')&&(a!='=')&&(a!='<'));}
inline void up(int&a,int b){if(a<b)a=b;}
inline void add(int*b,int x,int y){for(;x<=N;x+=x&-x)up(b[x],y);}
inline int ask(int*b,int x){int t=0;for(;x;x-=x&-x)up(t,b[x]);return t;}
int main(){
  read(n),read(k);
  for(i=1;i<=n;i++)read(a[i]);
  for(i=1;i<k;i++)read(s[i]);read(s[0]);
  for(i=1;i<=n;i++){
    f[i]=e[a[i]];
    up(f[i],ask(bl,a[i]-1));
    up(f[i],ask(bg,N-a[i]));
    f[i]++;
    if(s[f[i]%k]=='=')up(e[a[i]],f[i]);
    if(s[f[i]%k]=='<')add(bl,a[i],f[i]);
    if(s[f[i]%k]=='>')add(bg,N-a[i]+1,f[i]);
    up(ans,f[i]);
  }
  return printf("%d",ans),0;
}