[JSOI2008]火星人

题干：给你一段区间，并给你m个操作，操作包含：

1.计算以两点为起点的最长相同串长度。

2.将某字符改为另一字符。

3.在某位置后加入另一字符。

splay的题，还要配上hash，套上log级的验证，时间O(n*log^2);

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100500
#define seed 2333
#define ull unsigned long long
char ch[N],c[2];
int n,m,rt,tot,a[N];
ull mi[2*N];
struct Splay
{
    int ch[2],fa,vl,siz;
    ull hs;
}tr[N];
void update(int u)
{
    int l = tr[u].ch[0],r = tr[u].ch[1];
    tr[u].siz = tr[l].siz+tr[r].siz+1;
    tr[u].hs = tr[l].hs*mi[tr[r].siz+1]+tr[u].vl*mi[tr[r].siz]+tr[r].hs;
}
void rotate(int x)
{
    int y = tr[x].fa,z = tr[y].fa,k=(tr[y].ch[1]==x);
    tr[y].ch[k]=tr[x].ch[k^1];tr[tr[x].ch[k^1]].fa=y;
    tr[x].ch[k^1]=y;tr[y].fa=x;
    tr[z].ch[tr[z].ch[1]==y]=x;tr[x].fa=z;
    update(y);update(x);
}
void splay(int x,int goal)
{
    while(tr[x].fa!=goal)
    {
        int y = tr[x].fa,z = tr[y].fa;
        if(z!=goal)
            ((tr[y].ch[1]==x)^(tr[z].ch[1]==y))?rotate(x):rotate(y);
        rotate(x);
    }
    if(!goal)rt=x;
}
void build(int l,int r,int f)
{
    if(l>r)return ;
    int mid = (l+r)>>1;
    if(l==r)
    {
        tr[mid].hs = a[l];
    }else
    {
        build(l,mid-1,mid);
        build(mid+1,r,mid);
    }
    tr[mid].vl = a[mid];
    tr[mid].fa = f;
    tr[f].ch[mid>=f]=mid;
    update(mid);
}
int find(int x,int k)
{
    int t = tr[tr[x].ch[0]].siz;
    if(t>=k)return find(tr[x].ch[0],k);
    else if(k==t+1)return x;
    else return find(tr[x].ch[1],k-1-t);
}
int deal(int x,int y)
{
    x = find(rt,x),y = find(rt,y);
    splay(x,0);splay(y,x);
    return tr[y].ch[0];
}
ull query(int x,int y)
{
    x = deal(x,y+1);
    return tr[x].hs;
}
int get(int x,int y)
{
    int ans = 0;
    for(int i=20;i>=0;i--)
    {
        if(y+(1<<i)>n+1||x+(1<<i)>n+1)continue;
        if(query(x,x+(1<<i))==query(y,y+(1<<i)))
            x+=(1<<i),y+=(1<<i),ans|=(1<<i);
    }
    return ans;
}
void cg(int x,int k)
{
    x = find(rt,x);
    splay(x,0);
    tr[x].vl = k;
    update(x);
}
void insert(int x,int k)
{
    int l = find(rt,x),r = find(rt,x+1);
    splay(l,0);splay(r,l);
    tot++;
    tr[tot].vl = tr[tot].hs = k;
    tr[tot].fa = r;
    tr[r].ch[0]= tot;
    tr[tot].siz=1;
    update(r);
    update(rt);
}
int main()
{
    scanf("%s%d",ch+1,&m);
    n = strlen(ch+1);
    mi[0]=1;
    for(int i=1;i<=200000;i++)mi[i]=mi[i-1]*seed;
    for(int i=1;i<=n;i++)a[i+1]=ch[i]-'a'+1;
    build(1,n+2,0);
    rt=(3+n)>>1;
    tot = n+2;
    int x,y;
    for(int i=1;i<=m;i++)
    {
        scanf("%s%d",c,&x);
        if(c[0]=='Q')
        {
            scanf("%d",&y);
            printf("%d\n",get(x,y));
        }else if(c[0]=='R')
        {
            scanf("%s",c);
            cg(x+1,c[0]-'a'+1);
        }else
        {
            scanf("%s",c);
            n++;
            insert(x+1,c[0]-'a'+1);
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/LiGuanlin1124/p/9608700.html