标签：des style blog color io os ar for 数据
Description
Input
Output
Sample Input
3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 3 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA 3 CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalities AGATAC CATCATCAT
题意：先输入的测试数据的组数，接下来是每组数据的字符串数，每组字符串有60个字符，求最长公共子串，如果最长的公共子串长度小于3，就输出
no significant commonalities，否则输出字典序最小的最长公共子串。
思路：枚举第一个串的子串为模板与其他的字串比较。
ac代码：
1 #include<stdio.h> 2 #include<string.h> 3 char s1[200],s2[15][200]; 4 int next[200],cnt,n1,n2,n,ma; 5 void nextshuzu() //next数组 6 { 7 int i=0,j=-1; 8 next[0]=-1; 9 while(i<n1) 10 { 11 if(j==-1||s1[i]==s1[j]) 12 next[++i]=++j; 13 else 14 j=next[j]; 15 } 16 } 17 void kmp() 18 { 19 nextshuzu(); 20 int i=0,j=0; 21 ma=100; 22 for(int k=1; k<n; k++) 23 { 24 int m=0; 25 while(j<n2&&i<60) 26 { 27 if(j==-1||s1[j]==s2[k][i]) 28 { 29 i++; 30 j++; 31 } 32 else 33 j=next[j]; 34 if(j>m) //求出这个子串的最大值 35 m=j; 36 } 37 if(m<ma) //取出最小的那个，最小的都满足 38 ma=m; 39 } 40 } 41 int main() 42 { 43 int b; 44 char s[200]; 45 scanf("%d",&b); 46 while(b--) 47 { 48 scanf("%d",&n); 49 for(int i=0;i<n;i++) 50 scanf("%s",s2[i]); 51 int ans=0; 52 for(int i=0;i<58;i++) 53 { 54 strcpy(s1,s2[0]+i); 55 n2=60-i; 56 kmp(); 57 if(ans<ma) 58 { 59 ans=ma; 60 strncpy(s,s2[0]+i,ans); 61 s[ans]=‘\0‘; 62 } 63 else if(ma==ans) //开始第一次没写这里，wr后看别人的博客加上的，首先谢谢那个大神，在这我就先用了啊 64 { //如果长度相同，用字典序最小的那个 65 char s3[200]; 66 strncpy(s3,s2[0]+i,ans); 67 s3[ans]=‘\0‘; 68 if(strcmp(s,s3)>0) 69 strcpy(s,s3); 70 } 71 } 72 if(ans>2) 73 printf("%s\n",s); 74 else 75 printf("no significant commonalities\n"); 76 } 77 }
标签：des style blog color io os ar for 数据
原文地址：http://www.cnblogs.com/Xacm/p/3982309.html
