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;}