字符串专题：G - Milking Grid (二维kmp)

#include <stdio.h>
#include <string.h>
#define MAX 10010
#define MIN 110
//题意：给定一个n*m的字符矩阵，每行字符会有若干个相同的串组成，最后一个重复串不一定要完整。问这个矩阵可由面积最小为多少的矩阵组成。

int len,next[MAX];
int row[MAX],col[MIN];
char str[MAX][MIN],rostr[MIN][MAX];

int Kmp(char *str) {

int i,j,len;
len = strlen(str);
i = 0,j = -1;
next[i] = -1;
while (i < len) {
if (j == -1 || str[i] == str[j])
i++,j++,next[i] = j;
else j = next[j];
}
return len - next[len];//公共串长度,(最后一个重复串不一定要完整)
}

int gcd(int a,int b){
return b==0 ? a: gcd(b,a%b);
}

int lcm(int n,int m) {

return n / gcd(n,m) * m;
}

int main()
{
int i,j,k,m,n;
int r,c;

while (scanf("%d%d",&n,&m) != EOF) {

r = c = 1;
for (i = 0; i < n; ++i) {
scanf("%s",str[i]);
row[i] = Kmp(str[i]);
}
for (i = 0; i < n; ++i)
for (j = 0; j < m; ++j)
rostr[j][i] = str[i][j];//倒转一下矩阵
for (j = 0; j < m; ++j)
col[j] = Kmp(rostr[j]);

for (i = 0; i < m; ++i)
r = lcm(r,col[i]);//先由列判行，再由行判列（1 <= n <= 10,000 ,1 <= m <= 75）
r = n < r ? n : r;
for (i = 0; i < r; ++i)//最小重复矩阵一定在左上角，且不会超过r行。
c = lcm(c,row[i]);
c = m < c ? m : c;
printf("%d\n",r * c);
}
return 0;
}

/*
Sample Input

5 15
ABCDABCDABCDABC
BBABBABBABBABBA
ABCDABCDABCDABC
BBABBABBABBABBA
ABCDABCDABCDABC

4 12
ABABABABABAB
BBABBABBABBA
ABABABABABAB
BBABBABBABBA

Sample Output

24
最小矩阵为：
ABCDABCDABCD
BBABBABBABBA

col[0]:len-next[len]=5-3=2
col[1]:len-next[len]=5-4=1
c=lcm(col)=2*1=2;
row[0]:len-next[len]=15-11=4;
row[1]:len-next[len]=15-12=3;
r=lcm(row)=4*3=12;

12
最小矩阵为：
ABABAB
BBABBA

*/