POJ2185Milking Grid(最小覆盖子串 + 二维KMP)
题意: 一个r*c的矩形,求一个子矩形通过平移复制能覆盖整个矩形
关于一个字符串的最小覆盖子串可以看这里http://blog.csdn.net/fjsd155/article/details/6866991
把他分成对行和对列,对行覆盖最小就是n - next[n] ,然后求最小公倍数
对列的也是n - next[n], 然后求最小公倍数
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int Max = 100000 + 10; char str[Max][100]; int Next[Max]; int r, c; int getNextc(int n) { int k = -1; Next[0] = -1; int i = 0; while (i < c) { while (k != -1 && str[n][i] != str[n][k]) k = Next[k]; Next[++i] = ++k; } return c - Next[c]; // 是 c - Next[c],next[c]才是整个串 最长(前缀 == 后缀) } int getNextr(int n) { int k = -1; Next[0] = -1; int i = 0; while (i < r) { while (k != -1 && str[i][n] != str[k][n]) k = Next[k]; Next[++i] = ++k; } return r - Next[r]; } int gcd(int a, int b) { if (a == 0) return b; return gcd(a % b, a); } int getNum(int a, int b) { if (a > b) swap(a, b); int d = gcd(a, b); return a / d * b; } int main() { while (scanf("%d%d", &r, &c) != EOF) { getchar(); for (int i = 0; i < r; i++) scanf("%s", str[i]); int ans1 = 1, ans2 = 1; for (int i = 0; i < r; i++) // 按行处理 { memset(Next, 0, sizeof(Next)); ans1 = getNum(ans1, getNextc(i)); if (ans1 >= c) { ans1 = c; break; } } for (int i = 0; i < c; i++) { memset(Next, 0, sizeof(Next)); ans2 = getNum(ans2, getNextr(i)); if (ans2 >= r) { ans2 = r; break; } } printf("%d\n", ans1 * ans2); } return 0; }