[KMP或者暴力]POJ 3450 Corporate Identity

来源：互联网发布：淘宝开店创业计划书编辑：程序博客网时间：2024/05/21 23:34

传送门：http://poj.org/problem?id=3450

题目大意：前面那道题类似，求多个字符串的最长且字典序最小的公共子串，还是枚举子串，然后拿去和剩余主串匹配，保存最优解。

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN = 222,MAXM = 4444;char s[MAXM][MAXN],temp[MAXN],res[MAXN];int next[MAXN],n;void makenext(char *str){    int i = 0,j = -1,M = strlen(str);    next[0] = -1;    while(i<M){        if(str[i]==str[j]||j==-1)next[++i] = ++j;        else j = next[j];    }}int KMP(char *T,char *P){    int N = strlen(T),M = strlen(P),i = 0,j = 0;    while(i<N&&j<M){        if(T[i]==P[j]||j==-1)i++,j++;        else j = next[j];    }    return j==M?i-M:-1;}bool great(char *s1,char *s2){    int l1 = strlen(s1),l2 = strlen(s2);    if(l1!=l2)return l1>l2;    return strcmp(s1,s2)<0;}int main(){    while(scanf("%d",&n),n){        for(int i=0;i<n;i++)scanf("%s",s[i]);        int l = strlen(s[0]);res[0] = 0;        for(int i=0;i<l;i++){            for(int j=i;j<l;j++){                for(int k=0;k<j-i+1;k++){                    temp[k] = s[0][k+i];                }                temp[j-i+1] = 0;                makenext(temp);                int ok = 1;                for(int k=1;k<n;k++){                    if(KMP(s[k],temp)==-1){                        ok = 0;break;                    }                }                //cout<<temp<<" "<<ok<<endl;                if(ok&&great(temp,res)){                    strcpy(res,temp);                }            }        }        if(res[0]){            printf("%s\n",res);        }else{            printf("IDENTITY LOST\n");        }    }    return 0;}