POJ3080 Blue Jeans 【KMP 暴力水过】

题目描述

求n个字符串的最长公共序列，若长度相同，输出字典序最小的。若长度小于3，输出no significant commonalities

Sample Input

3

2

GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

3

GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA

GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA

GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA

3

CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalities

AGATAC

CATCATCAT

解题思路

暴力0ms可过....用ans储存当前最优解。

下面是代码

#include <cstdio>
#include <cstring>
const int maxn = 61;
int n;
int anslen = 0;
char ans[maxn];
char s[4010][maxn];
char p[maxn];//模式串
int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        anslen = 0;
        for(int i = 0 ; i < n ; i ++) scanf("%s",s[i]);
        int len1 = strlen(s[0]);
        for(int i = 0 ; i < len1 ; i ++) {
            for(int j = i ; j < len1 ; j ++) {
                /** 模式串是s[0][i,i+1,...,j] */
                for(int k = i ; k <= j ; k ++) {
                    p[k-i] = s[0][k];
                }
                p[j-i+1] = '\0';
                /** 模式串已就绪 */
                bool flag = true;//记录其他主串是否都有p串
                int len = j-i+1;//长度
                if(len < anslen) continue;
                for(int k = 1 ; k < n ; k ++) {
                    if(strstr(s[k],p) == NULL) {
                        flag = false;
                        break;
                    }
                }
                if(flag) {
                    if(len == anslen) {//看字典序
                        if(strcmp(p,ans) < 0) {
                            strcpy(ans,p);
                        }
                    }else {
                        strcpy(ans,p);
                        anslen = len;
                    }
                }
            }
        }
        if(anslen < 3) printf("no significant commonalities\n");
        else printf("%s\n",ans);
    }
    return 0;
}