BZOJ3277 串 【后缀数组】【二分答案】【主席树】

题目分析：

用"$"连接后缀数组，然后做一个主席树求区间内不同的数的个数。二分一个前缀长度再在主席树上求不同的数的个数。

代码:

#include<bits/stdc++.h>
using namespace std;

const int maxn = 132000;
const int N = 130000;

int m,n,k;
string hmk[maxn],str;
int sum[maxn];

int sa[maxn],rk[maxn],X[maxn],Y[maxn];
int height[maxn],h[maxn],RMQ[maxn][20],Tnum;
long long ans[maxn];

struct node{
    int ch[2],sum;
}CMT[maxn*35];

int chk(int x,int k){
    return rk[sa[x]]==rk[sa[x-1]]&&rk[sa[x]+(1<<k)]==rk[sa[x-1]+(1<<k)];
}

void getsa(){
    for(int i=0;i<n;i++) X[str[i]]++;
    for(int i=1;i<=N;i++) X[i] += X[i-1];
    for(int i=n-1;i>=0;i--) sa[X[str[i]]--] = i;
    for(int i = 2, num = 1;i <= n;i++)
    rk[sa[i]] = (str[sa[i]] == str[sa[i-1]]?num:++num);
    rk[sa[1]] = 1;
    for(int k=1;(1<<k-1)<=n;k++){
    for(int i=1;i<=N;i++) X[i] = 0;
    for(int i=n-(1<<k-1);i<n;i++) Y[i-n+(1<<k-1)+1]=i;
    for(int i=1,j=(1<<k-1)+1;i<=n;i++)
        if(sa[i]>=(1<<k-1))Y[j++]=sa[i]-(1<<k-1);
    for(int i=0;i<n;i++) X[rk[i]]++;
    for(int i=1;i<=N;i++) X[i]+=X[i-1];
    for(int i=n;i>=1;i--) sa[X[rk[Y[i]]]--] = Y[i];
    int num = 1; Y[sa[1]] = 1;
    for(int i=2;i<=n;i++) Y[sa[i]] = (chk(i,k-1)?num:++num);
    for(int i=0;i<n;i++) rk[i] = Y[i];
    if(num == n) break;
    }
}
void getheight(){
    for(int i=0;i<n;i++){
    if(i) h[i] = max(0,h[i-1]-1); else h[i] = 0;
    if(rk[i] == 1) continue;
    int comp = sa[rk[i]-1];
    while(str[comp+h[i]] == str[i+h[i]])h[i]++;
    }
    for(int i=0;i<n;i++) height[rk[i]] = h[i];
    for(int i=1;i<=n;i++) RMQ[i][0] = height[i];
    for(int k=1;(1<<k)<=n;k++){
    for(int i=1;i<=n;i++){
        if(i+(1<<k-1)>n) RMQ[i][k]=RMQ[i][k-1];
        else RMQ[i][k] = min(RMQ[i][k-1],RMQ[i+(1<<k-1)][k-1]);
    }
    }
}

int pww[maxn];
int getLCP(int L,int R){
    if(L == R) return n-sa[L];
    L++;
    int k = pww[R-L+1];
    return min(RMQ[L][k],RMQ[R-(1<<k)+1][k]);
}

void build_empty_tree(int now,int tl,int tr){
    if(tl == tr) return;
    int mid = (tl+tr)/2;
    Tnum++; CMT[now].ch[0] = Tnum;
    build_empty_tree(Tnum,tl,mid);
    Tnum++; CMT[now].ch[1] = Tnum;
    build_empty_tree(Tnum,mid+1,tr);
}

void Modify(int lst,int now,int tl,int tr,int pos,int dt){
    CMT[now] = CMT[lst];
    if(tl == tr){CMT[now].sum += dt;}
    else{
    int mid = (tl+tr)/2;
    if(pos <= mid){
        CMT[now].ch[0] = ++Tnum;
        Modify(CMT[lst].ch[0],Tnum,tl,mid,pos,dt);
    }else{
        CMT[now].ch[1] = ++Tnum;
        Modify(CMT[lst].ch[1],Tnum,mid+1,tr,pos,dt);
    }
    CMT[now].sum += dt;
    }
}

int imp[maxn],his[maxn];
void build_CMT(){
    his[0] = 1;
    for(int i=1;i<=n;i++){
    if(imp[sum[sa[i]]]){
        int z = ++Tnum;
        Modify(his[i-1],z,1,n,imp[sum[sa[i]]],-1);
        his[i] = ++Tnum; imp[sum[sa[i]]] = i;
        Modify(z,his[i],1,n,i,1);
    }else{
        his[i] = ++Tnum; imp[sum[sa[i]]] = i;
        Modify(his[i-1],his[i],1,n,i,1);
    }
    }
}

int query(int now,int tl,int tr,int l,int r){
    if(tl >= l && tr <= r) return CMT[now].sum;
    if(tl > r || tr < l) return 0;
    int mid = (tl+tr)/2;
    int as=query(CMT[now].ch[0],tl,mid,l,r)+query(CMT[now].ch[1],mid+1,tr,l,r);
    return as;
}

int rgt[maxn];
void work(){
    getsa();
    getheight();
    sum[n] = 1;rgt[n] = n;
    for(int i=n-1;i>=0;i--){
    sum[i] = sum[i+1]+(str[i] == '$');
    if(str[i] == '$') rgt[i] = i;
    else rgt[i] = rgt[i+1];
    }
    Tnum = 1;
    build_empty_tree(1,1,n);
    build_CMT();
    for(int i=m;i<=n;i++){
    int l = 0,r = rgt[sa[i]]-sa[i];
    while(l < r){
        int mid = (l+r+1)/2;
        int al = 1,ar = i;
        while(al < ar){
        int midd = (al+ar)/2;
        if(getLCP(midd,i) >= mid) ar = midd;
        else al = midd+1;
        }
        int tl  = i,tr = n;
        while(tl < tr){
        int midd = (tl+tr+1)/2;
        if(getLCP(i,midd) >= mid) tl = midd;
        else tr = midd-1;
        }
        int mmp = query(his[tl],1,n,al,tl);
        if(mmp >= k) l = mid;
        else r = mid-1;
    }
    ans[m-sum[sa[i]]+1] += l;
    }
    for(int i=1;i<=m;i++) printf("%lld ",ans[i]);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> m >> k;
    for(int i=1;i<=m;i++) cin >> hmk[i];
    for(int i=1;i<m;i++) str += hmk[i],str += '$';
    str += hmk[m];
    n = str.length();
    pww[1] = 0;
    for(int i=2;i<=n;i++) pww[i] = pww[i/2]+1;
    work();
    return 0;
}