【洛谷 P4248】 [AHOI2013]差异（后缀自动机）

题目链接

\[ans=\sum_{1<=i<j<=n}len(T_i)+len(T_j)-2*lcp(T_i,T_j) \]

观察这个式子可以发现，前面两个\(len\)是常数，后面的其实就是反串有每对前缀的相同后缀乘以其长度之和。
两个前缀的相同后缀就是这两个串在parent tree上对应的点的\(LCA\)，于是直接树上统计就行了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1000010;
struct SAM{
    int ch[26];
    int len, fa;
}sam[MAXN << 1];
int las = 1, cnt = 1, f[MAXN << 1];
struct Edge{
    int next, to;
}e[MAXN << 1];
int head[MAXN << 1], num, n;
inline void Add(int from, int to){
    e[++num].to = to; e[num].next = head[from]; head[from] = num;
}
inline void add(int c){
    int p = las; int np = las = ++cnt;
    sam[np].len = sam[p].len + 1; f[cnt] = 1;
    for(; p && !sam[p].ch[c]; p = sam[p].fa) sam[p].ch[c] = np;
    if(!p) sam[np].fa = 1;
    else{
        int q = sam[p].ch[c];
        if(sam[q].len == sam[p].len + 1) sam[np].fa = q;
        else{
            int nq = ++cnt; sam[nq] = sam[q];
            sam[nq].len = sam[p].len + 1;
            sam[q].fa = sam[np].fa = nq;
            for(; p && sam[p].ch[c] == q; p = sam[p].fa) sam[p].ch[c] = nq;
        }
    }
}
char a[MAXN];
long long ans;
void dfs(int u){
    long long tmp = 0;
    for(int i = head[u]; i; i = e[i].next){
        dfs(e[i].to);
        tmp += (long long)f[u] * f[e[i].to];
        f[u] += f[e[i].to];
    }
    ans += (long long)tmp * 2 * sam[u].len;
}
int main(){
    scanf("%s", a + 1);
    n = strlen(a + 1);
    for(int i = 1; i <= n; ++i)
       add(a[i] - 'a');
    for(int i = 2; i <= cnt; ++i)
        Add(sam[i].fa, i);
    dfs(1);
    printf("%lld\n", (long long)n * (n - 1) * (n + 1) / 2 - ans);
    return 0;
}