hdu2328(暴力kmp）

這個題我想練一下kmp就沒用後綴數組做

後綴數組的思路也很簡單 鏈接所有串 然後二分答案就可以了

也可以暴力枚舉後綴 對每一個後綴用kmp找到最小公共前綴就可以

我算着複雜度是（200*4000*400）可能3s勉強能過...但是隻用了幾百ms 感覺數據出水了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
char s[4000+100][400];
int knext[4000+100];
void ekmp(char *s,int len){
    int i = 0;
    int j = -1;
    knext[0] = -1;
    while(i<len){
        while(j!=-1&&s[i]!=s[j])j = knext[j];
        knext[++i] = ++j;
    }
}
int kmp_count(char *x,int n,char *y,int m){
    int i,j;
    i = 0;
    j = 0;
    ekmp(x,n);
    int maxs = 0;
    while(i<m){
        while(j!=-1&&x[j]!=y[i])j = knext[j];
        ++i;++j;
        if(j>=n){
            return n;
        }
        maxs = max(maxs,j);
    }
    return maxs;
}
char ss[20000];
int lens[4000+100];
char res[20000];
int main(){
    int n;
    while(scanf("%d",&n)&&n){
        int minn = 0x3f3f3f3f;
        int k = -1;
        for(int i = 0;i<n;++i){
            scanf("%s",s[i]);
            lens[i] = strlen(s[i]);
            if(lens[i]<minn){
                minn = lens[i];
                k = i;
            }

        }
        int maxs = 0;
        int len2 = lens[k];
        int st = 0;
        int now_st = 0;
        for(int j = 0;j<len2;++j){

                st = j;

                int cnt = 0;
                for(int i = j;i<len2;++i){
                    ss[cnt++] = s[k][i];
                }
                ss[cnt] = '\0';
                int len = cnt;
                int minn1 = 0x3f3f3f3f;
                for(int i = 0;i<n;++i){
                    int len1 = lens[i];
                    int index = kmp_count(ss,len,s[i],len1);
                    minn1 = min(minn1,index);
                }
                if(minn1>maxs){
                maxs = minn1;
                for(int h = 0;h<maxs;++h){
                    res[h] = ss[h];
                }
                res[maxs] = '\0';
            }
            else if(minn1==maxs&&strcmp(res,ss)>0){
                    for(int h = 0;h<maxs;++h){
                        res[h] = ss[h];
                    }
                    res[maxs] = '\0';
            }

        }
        if(maxs){
            printf("%s\n",res);
        }
        else{
            cout<<"IDENTITY LOST"<<endl;
        }


    }

}
/*
 3
 abba
 abcd
  cdab
 *
 */