题解 P3181 [HAOI2016]找相同字符

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题面

直接对两个串建广义sam
然后分别搞出A,B的right集合大小
然后乘起来再求和就可以了

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#include <bits/stdc++.h>
using namespace std;
int buc[800010], a[800010];
namespace sam {
	int son[800010][26], par[800010], cnt, last, len[800010], sze[800010][2];
	inline void clear() {
		cnt = last = 1;
	}
	inline void add(int x) {
		int p = last;
		int np = last = ++cnt;
		len[np] = len[p] + 1;
		for(; p && son[p][x] == 0; p = par[p]) son[p][x] = np;
		if(p == 0) par[np] = 1;
		else {
			int q = son[p][x];
			if(len[q] == len[p] + 1) par[np] = q;
			else {
				int nq = ++cnt;
				par[nq] = par[q];
				memcpy(son[nq], son[q], sizeof(int[26]));
				len[nq] = len[p] + 1;
				par[np] = par[q] = nq;
				for(; son[p][x] == q; p = par[p]) son[p][x] = nq;
			}
		}
	}
	inline void get(char *s, int len, int id) {
		for(int i = 0, now = 1; i < len; i++) sze[now = son[now][s[i] - 'a']][id] = 1;
	}
	inline long long calc() {
		for(int i = 1; i <= cnt; i++) buc[len[i]] ++;
		for(int i = 1; i <= cnt; i++) buc[i] += buc[i - 1];
		for(int i = 1; i <= cnt; i++) a[buc[len[i]]--] = i;
		for(int i = cnt; i > 0; i--) {
			sze[par[a[i]]][0] += sze[a[i]][0];
			sze[par[a[i]]][1] += sze[a[i]][1];
		}
		long long ans = 0;
		for(int i = 1; i <= cnt; i++) ans += 1ll * sze[i][0] * sze[i][1] * (len[i] - len[par[i]]);
		return ans;
	}
};
using namespace sam;
char A[200010], B[200010];
int main() {
	clear();
	int lena, lenb;
	scanf("%s%s", A, B);
	lena = strlen(A), lenb = strlen(B);
	for(int i = 0; i < lena; i++) add(A[i] - 'a');
	last = 1;
	for(int i = 0; i < lenb; i++) add(B[i] - 'a');
	get(A, lena, 0);
	get(B, lenb, 1);
	return cout << calc() << endl, 0;
}