poj 2185 Milking Grid(KMP)

题目链接：poj 2185 Milking Grid

题目大意：给定一个N?M的矩阵，找到一个最小的r?c的矩阵，使得原矩阵可以由若干个小矩阵组成，输出r?c的值。

解题思路：将行和列分别看成字符串，然后hash，对hash后的数组分别求KMP。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef unsigned long long ll;
const ll x = 123;
const int maxr = 10005;
const int maxc = 80;

int N, M, jump[maxr];
ll r[maxr], c[maxc];
char s[maxr][maxc];

void init () {
    for (int i = 1; i <= N; i++)
        scanf("%s", s[i] + 1);
    memset(r, 0, sizeof(r));
    memset(c, 0, sizeof(c));

    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= M; j++)
            r[i] = r[i] * x + s[i][j];
    }

    for (int j = 1; j <= M; j++) {
        for (int i = 1; i <= N; i++)
            c[j] = c[j] * x + s[i][j];
    }
}

int kmp (int n, ll* a) {
    int p = 0;
    for (int i = 2; i <= n; i++) {
        while(p && a[p+1] != a[i])
            p = jump[p];

        if (a[p+1] == a[i])
            p++;
        jump[i] = p;
    }
    return n - jump[n];
}

int solve () {
    return kmp(N, r) * kmp(M, c);
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        init();
        printf("%d\n", solve());
    }
    return 0;
}