poj 2185

http://poj.org/problem?id=2185

题意：求最小的模式块，使其无限扩展后包含给你的矩阵块（看别人题解才懂的题意）；

分析：假设存在一个模式块可以满足上述条件，那么必然存在一个起点在（0,0）的模式块满足上述条件；

　　对于每一行，我们找出所有可以满足条件的前缀记录下长度，那么满足所有行的最短的长度就是该模式块的宽r；

　　对于模式块的长，我们把宽r的字符串压缩看出一个字符，然后再进行KMP，找出该字符串的最小循环串，即长l；

答案就是r * l;

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;
const int N = 10000+10;
int f[N];
void getFail(char *s,int f[]) {
    int m = strlen(s);
    f[0] = -1;
    for (int i = 1; i < m; i++) {
        int j = f[i-1];
        while (j != -1 && s[j+1] != s[i]) j = f[j];
        if (s[j+1] == s[i]) j++;
        f[i] = j;
    }
 //   for (int i = 0; i < m; i++) cout<<f[i]<<" ";cout<<endl;
}
int n,m;
char mz[N][80];
int cnt[N];
int check(char *s,int x) {
    int j = 0;
    for (int i = 0; i < m; i++) {
        if (s[i] != s[j]) return 0;
        j++;
        if (j > x) j = 0;
    }
    return 1;
}
void solve(int r){
    f[0] = -1;
    for (int i = 1; i < n; i++) {
        int j = f[i-1];
        while (j != -1 && strcmp(mz[j+1],mz[i])) j = f[j];
        if (strcmp(mz[j+1],mz[i]) == 0) j++;
        f[i] = j;
    }
   // for (int i = 0; i < n; i++) cout<<f[i]<<" ";cout<<endl;
    printf("%d\n",(n - 1 - f[n-1]) * (r));
}
int main(){
 //   getFail("ababab",f);
    while (~scanf("%d%d",&n,&m)) {
        memset(cnt,0,sizeof(cnt));

        for (int i = 0; i < n; i++) {
            scanf("%s",mz[i]);
            getFail(mz[i],f);
            int j = m-1;
            while (j != -1) {
              //  cout<<m - 1 - f[j]<<endl;
                cnt[m - 1 - f[j]]++;
                j = f[j];
            }
        }
        int r = 0;
        for (int i = 1; i <= m; i++) if (cnt[i] == n) {
            r = i; break;
        }
        for (int i = 0; i < n; i++) {
            mz[i][r] = 0;
        }

        solve(r);
    }
    return 0;
}