H 洪尼玛的保险箱

题意：

求两个字符串的公共子串个数，且该公共子串在第一个串中以奇数位置结尾，在第二个串中以偶数位置结尾

思路：

建广义后缀自动机后根据限制条件求出每个状态的\(|endpos|\)，统计答案即可

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<vector>
using namespace std;
typedef long long ll;
const int Maxn = 400010;
const int N = 2;
struct Suffix_Automata{
    int maxlen[Maxn], trans[Maxn][26], link[Maxn], Size;
    int sz[Maxn][N];
	void clr(int x){
        maxlen[x] = link[x] = 0;
        memset(trans[x], 0, sizeof(trans[x]));
        memset(sz[x], 0, sizeof(sz[x]));
    }
    void init(){
        Size = 1;
        clr(1);
    }
    int insert(int ch, int last, int id, int idx){
        if (trans[last][ch]){
            int p = last, x = trans[p][ch];
            if (maxlen[p] + 1 == maxlen[x]){
                if(id ^ idx)
                    sz[x][id] = 1; 
                return x;
            }else{
                int y = ++Size;
                clr(y);
                maxlen[y] = maxlen[p] + 1;
                for (int i = 0; i < 26; ++i)
                    trans[y][i] = trans[x][i];
                while (p && trans[p][ch] == x)
                    trans[p][ch] = y, p = link[p];
                link[y] = link[x], link[x] = y;
                if(id ^ idx)
                    sz[y][id] = 1;
                return y;
            }
        }
        int z = ++Size, p = last;
        clr(z);
        if(id ^ idx)
            sz[z][id] = 1;
        maxlen[z] = maxlen[last] + 1;
        while (p && !trans[p][ch])
            trans[p][ch] = z, p = link[p];
        if (!p) link[z] = 1;
        else{
            int x = trans[p][ch];
            if (maxlen[p] + 1 == maxlen[x]) link[z] = x;
            else{
                int y = ++Size;
                clr(y);
                maxlen[y] = maxlen[p] + 1;
                for (int i = 0; i < 26; ++i)
                    trans[y][i] = trans[x][i];
                while (p && trans[p][ch] == x)
                    trans[p][ch] = y, p = link[p];
                link[y] = link[x], link[z] = link[x] = y;
            }
        }
        return z;
    }
    int c[Maxn], a[Maxn];
    void rsort(int n){
        for(int i = 0; i <= n; ++i) c[i] = a[i] = 0;;
        for (int i = 1; i <= Size; i++) c[maxlen[i]]++;
        for (int i = 1; i <= n; i++) c[i] += c[i - 1];
        for (int i = 1; i <= Size; i++) a[c[maxlen[i]]--] = i;
        for (int i = Size; i >= 1; i--) sz[link[a[i]]][0] += sz[a[i]][0], sz[link[a[i]]][1] += sz[a[i]][1];
    }
    void solve(){
        ll ans = 0;
        for(int i = 1; i <= Size; ++i)
            ans += 1ll * sz[i][0] * sz[i][1] * (maxlen[i] - maxlen[link[i]]);
        printf("%lld\n", ans);

    }
}sam;
char s1[100010], s2[100010];
int main(){
    while(~scanf("%s%s", s1 + 1, s2 + 1)){
        int n = strlen(s1 + 1), m = strlen(s2 + 1);
        sam.init();
        int last = 1;
        for(int i = 1; i <= n; ++i) last = sam.insert(s1[i] - 'a', last, 0, i & 1);
        last = 1;
        for(int i = 1; i <= m; ++i) last = sam.insert(s2[i] - 'a', last, 1, i & 1);
        sam.rsort(max(n, m));
        sam.solve();
    }
    return 0;
}