poj2185 (kmp algorithm next array seek minimum cycle section, thinking)

Topic link: https: //vjudge.net/problem/POJ-2185

The meaning of problems: r × c matrix given by the capital letters, seeking a minimal matrix such that the sub-matrix can be composed of a large matrix, but does not require large matrix consisting of just a small matrix, i.e., the boundary portion will be left vacant (see examples) .

Ideas:
　　each row treated as a character, and then find next row array of r, then r-nex [r] is the minimum cycle in a row unit section size ans1.

　　The treated as a character for each column, and then to seek next column array c, the c-nex [c] is the minimum size of the cell cycle as a section ans2.

　　That is the final answer ans1 * ans2. (Area of ​​the sub-matrix)

AC Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxr=10005;
const int maxc=80;
char s[maxr][maxc];
int ans1,ans2,r,c,nex1[maxr],nex2[maxc];

bool check1(int i,int j){
    for(int k=0;k<c;++k)
        if(s[i][k]!=s[j][k])
            return false;
    return true;
}

bool check2(int i,int j){
    for(int k=0;k<r;++k)
        if(s[k][i]!=s[k][j])
            return false;
    return true;
}

void get_next1(){
    int j;
    j=nex1[0]=-1;
    for(int i=1;i<r;++i){
        while(j>-1&&!check1(i,j+1)) j=nex1[j];
        if(check1(i,j+1)) ++j;
        nex1[i]=j;
    }
}

void get_next2(){
    int j;
    j=nex2[0]=-1;
    for(int i=1;i<c;++i){
        while(j>-1&&!check2(i,j+1)) j=nex2[j];
        if(check2(i,j+1)) ++j;
        nex2[i]=j;
    }
}

int main () {
    scanf("%d%d",&r,&c);
    for(int i=0;i<r;++i)
        scanf("%s",s[i]);
    get_next1();
    get_next2();
    ans1=r-(nex1[r-1]+1);
    ans2=c-(nex2[c-1]+1);
    printf("%d\n",ans1*ans2);
    return 0;
}