Samjia 和矩阵[loj6173](Hash+后缀数组)

传送门

本题要求本质不同的子矩阵，即位置不同也算相同(具体理解可以看样例自己yy)。

我们先看自己会什么，我们会求一个字符串中不同的子串的个数。我们考虑把子矩阵变成一个字符串。

先枚举矩阵的宽度，记为w(1<=w<=m)。再把一行之内的连续的w的字符用字符串hash哈成一个整数，再把这个整数hash成一个较小的数（相当于之前字符串的一个字符）。

把最后完成的矩阵在把没一列接起来，形成一个字符串(每列后要加一个字符如：1 2 3 4 2 3 3 5)。然后对这个串求我们会的子串个数(减去height值)即可。

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll N = 120, M = 120;
const ull P = 13331;
ll len, tot, ans;
ll n, m;
ll s[N * M], rnk[N * M], sa[N * M], sum[N * M], v1[N * M], v2[N * M], height[N * M];
char arr[N][M];
ull _hash[N][M];
map<ull, ll> mp;
bool cmp(ll *t, ll a, ll b, ll l) {
    return t[a] == t[b] && t[a + l] == t[b + l];
}
void da() {
    ll i, j, p = 0;
    for (i = 1; i <= tot; i++) sum[i] = 0;
    for (i = 1; i <= len; i++) sum[rnk[i] = s[i]]++;
    for (i = 2; i <= tot; i++) sum[i] += sum[i - 1];
    for (i = len; i >= 1; i--) sa[sum[rnk[i]]--] = i;
    for (j = 1; j <= len; j *= 2, tot = p) {
        for (p = 0, i = len - j + 1; i <= len; i++) v2[++p] = i;
        for (i = 1; i <= len; i++) if (sa[i] > j) v2[++p] = sa[i] - j;
        for (i = 1; i <= len; i++) v1[i] = rnk[v2[i]];
        for (i = 1; i <= tot; i++) sum[i] = 0;
        for (i = 1; i <= len; i++) sum[v1[i]]++;
        for (i = 2; i <= tot; i++) sum[i] += sum[i - 1];
        for (i = len; i >= 1; i--) sa[sum[v1[i]]--] = v2[i];
        for (swap(rnk, v2), rnk[sa[1]] = 1, i = 2, p = 2; i <= len; i++) {
            rnk[sa[i]] = cmp(v2, sa[i - 1], sa[i], j) ? p - 1 : p++;
        }
    } 
}
void calheight() {
    ll i, j, p = 0;
    for (i = 1; i <= len; i++) {
        if (p) p--;
        j = sa[rnk[i] - 1];
        while (s[i + p] == s[j + p]) p++;
        height[rnk[i]] = p;
    }
}
ull ksm[150];
int main() {
    ksm[0] = 1;
    for (int i = 1; i <= 130; i++) {
        ksm[i] = ksm[i - 1] * P;
    }
    scanf("%lld%lld", &n, &m);
    for (ll i = 1; i <= n; i++) {
        scanf("%s", arr[i] + 1);
        for (ll j = 1; j <= m; j++) {
            _hash[i][j] = _hash[i][j - 1] * P + arr[i][j] - 'A' + 1;
        }
    }
    for (ll w = 1; w <= m; w++) {
        tot = 0, len = 0;
        mp.clear();
        for (ll j = 1; j + w - 1 <= m; j++) {
            for (ll i = 1; i <= n; i++) {
                ull tmp = _hash[i][j + w - 1] - _hash[i][j - 1] * ksm[w];
                if (mp[tmp] == 0) {
                    mp[tmp] = ++tot;
                }
                s[++len] = mp[tmp];
            }
            s[++len] = ++tot;
        }
        da();
        calheight();
        ans += n * (n + 1) / 2 * (m - w + 1); 
        for (ll i = 2; i <= len; i++) {
            ans -= height[i];
        }
    }
    cout << ans;
    return 0;
}