Poi2010 Monotonicity 2

树状数组优化dp

可以证明最优解一定是通过之前的最优转移过来的，所以每一个点只需要保存以该节点为结尾的最长长度即可

对于不同符号，等于号维护数组，大于小于维护树状数组

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#define N 500005using namespace std;int n,m,a[N],f[N],ans,c[2][2*N],ff[2*N],maxn;char s[N];int lowbit(int x){    return x&(-x);}void update(int k,int x,int y){    while(x<=maxn){        c[k][x]=max(c[k][x],y);        x+=lowbit(x);    }}int query(int k,int x){    int ans=0;    while(x){        ans=max(c[k][x],ans);        x-=lowbit(x);    }    return ans;}int main(){    //freopen("mot.in","r",stdin);    //freopen("mot.out","w",stdout);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++){        scanf("%d",&a[i]);        maxn=max(maxn,a[i]);    }    for(int i=1;i<=m;i++){        s[i]=getchar();        while(s[i]!='<'&&s[i]!='>'&&s[i]!='=')        s[i]=getchar();    }    int x1,x2,x3; char ch;    for(int i=1;i<=n;i++){        x1=query(0,a[i]-1);        x2=query(1,maxn-a[i]);        x3=ff[a[i]];        f[i]=max(x1,max(x2,x3))+1;        ch=s[(f[i]-1)%m+1];        if(ch=='<') update(0,a[i],f[i]);        if(ch=='>') update(1,maxn-a[i]+1,f[i]);        if(ch=='=') ff[a[i]]=max(ff[a[i]],f[i]);    }    for(int i=1;i<=n;i++){        //printf("%d  %d\n",i,f[i]);        ans=max(ans,f[i]);    }    printf("%d\n",ans);    return 0;}