bzoj 3238 lcp

Description

Input

一行，一个字符串S

Output

 

一行，一个整数，表示所求值

Sample Input

cacao

Sample Output

54

这道题需要知道后缀自动机的一个性质，两个串的最长公共后缀，位于这两个串对应状态在parent树的最近公共祖先上。

分析一下题目中的式子，发现sigma len(Ti)+len(Tj)可以直接进行求解。关键是如何求解lcp ，因为上面的性质求得是最长公共后缀，所以我们将原串倒置，建立后缀自动机，得到的最长公共后缀就是原串的最长公共前缀。

知道这个后，问题就变成了，如何求解某个状态是多少对节点的后缀，这个可以用树形dp的思想，从下往上依次更新。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef pair<long long int,long long int> ii;typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=5e5+10;const int maxx=600005;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;int root,last;int tots;  //总结点int l;  //字符串长度int sv[maxn*2];int dp[maxn*2];char s[maxn];LL sum[maxn*2];struct sam_node{    int fa,son[26];    int len;    void init(int _len)    {        len = _len;        fa = -1;        memset(son,-1,sizeof(son));    }}t[maxn*2];void sam_init(){    tots = 0;    root = last = ++tots;    t[tots].init(0);}void extend(char ch){    int w=ch-'a';    int p=last;    int np=++tots;t[tots].init(t[p].len+1);    sv[l]=np;    int q,nq;    while(p!=-1&&t[p].son[w]==-1){t[p].son[w]=np;p=t[p].fa;}    if (p==-1) t[np].fa=root;    else    {        q=t[p].son[w];        if (t[p].len+1==t[q].len){t[np].fa=q;}        else        {            nq=++tots;t[nq].init(0);            t[nq]=t[q];            t[nq].len=t[p].len+1;            t[q].fa=nq;t[np].fa=nq;            while(p!=-1&&t[p].son[w]==q){t[p].son[w]=nq;p=t[p].fa;}        }    }    last=np;}int w[maxn*2], r[maxn*2];//w 一倍就够void topsort(){    for(int i = 0; i <= l; ++i) w[i] = 0;    for(int i = 1; i <= tots; ++i) w[t[i].len]++;    for(int i = 1; i <= l; ++i) w[i] += w[i-1];    for(int i = tots; i >= 1; --i) r[w[t[i].len]--] = i;    r[0] = 0;}LL ans=0;void DP(){    topsort();    //FOR(0,tots,i) dp[i]=0;    int now=root;    fOR(l-1,0,i)    {        now=t[now].son[s[i]-'a'];        dp[now]++;        sum[now]++;    }    for(int i=tots;i;i--)    {        int p=r[i];        if(t[p].fa != -1) dp[t[p].fa] += dp[p];        //cout<<dp[t[p].fa]<<endl;    }    for(int i=tots;i;i--)    {        int x=r[i];        if(t[x].fa != -1)        {            ans+=1ll*sum[t[x].fa]*dp[x]*t[t[x].fa].len;            sum[t[x].fa]+=dp[x];        }        //cout<<ans<<" "<<dp[x]<<" "<<sum[t[x].fa]<<endl;    }}int main(){    scanf("%s",s);    l=strlen(s);    sam_init();    for(int i=l-1;i>=0;i--)        extend(s[i]);    DP();    printf("%lld\n",1ll*l*(l+1)/2*(l-1)-ans*2);}