HDU 1238 Substrings(求公共正反向连续子串)

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Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8504 Accepted Submission(s): 3931

Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

Output

There should be one line per test case containing the length of the largest string found.

Sample Input

23ABCDBCDFFBRCD2roseorchid

Sample Output

Author

Asia 2002, Tehran (Iran), Preliminary

//优化 #include <iostream>#include <cstdio>#include <string>#include <string.h>using namespace std;const int N=100+10;int n,id,re,ilen;char c[N][N];int len[N];int main(){int t,i,small,j,k,z,zz;bool ok;scanf("%d",&t);while(t--){re=0;small=N;scanf("%d",&n);for(i=0;i<n;i++){   scanf("%s",c[i]);   len[i]=strlen(c[i]);   if(len[i]<small)     {     small=len[i];     id=i;}}ilen=len[id];for(i=ilen;re==0&&i>0;i--){   //枚举子串长度   for(j=0;re==0&&j+i-1<ilen;j++){   //枚举模板字串的起始点 for(k=0;k<n;k++){if(k==id)continue;ok=0;for(z=0;z+i-1<len[k];z++) {  //从左开始    for(zz=0;zz<i;zz++){      if(c[k][z+zz]!=c[id][j+zz])        break;   }   if(zz==i){   ok=1;   break;   }  }for(z=len[k]-1;!ok&&z-i+1>=0;z--) {  //从右开始    for(zz=0;zz<i;zz++){      if(c[k][z-zz]!=c[id][j+zz])        break;   }   if(zz==i){   ok=1;   break;   }  }  if(!ok)    break;}   if(k==n){  re=i;  break;  }  } }    printf("%d\n",re);}return 0;}/*哎呀   循环放错了   不过AC了 #include <iostream>#include <cstdio>#include <string>#include <string.h>#include <algorithm>using namespace std;const int N=100+10;int n,id,re,ilen,rlen;char c[N][N];int len[N];int main(){int t,i,small,j,k,z,zz;bool ok;scanf("%d",&t);while(t--){re=0;rlen=0;small=N;scanf("%d",&n);for(i=0;i<n;i++){   scanf("%s",c[i]);   len[i]=strlen(c[i]);   if(len[i]<small)     {     small=len[i];     id=i;}}ilen=len[id];for(i=0;i<ilen;i++){  //枚举模板字符串的字符起始位置   for(j=ilen-i;i+j-1<ilen&&j>0;j--) {    //子串的长度       for(k=0;k<n;k++){   //枚举每个字符串          if(k==id)        continue;        ok=0;for(z=0;z+j-1<len[k];z++){ //左开始的位置 for(zz=0;zz<j;zz++)  //每个字符比较 {    if(c[k][z+zz]!=c[id][i+zz])break;}  if(zz==j)   {      ok=1;  break;   }      }  for(z=len[k]-1;!ok&&z-j+1>=0;z--){  //从右开始   for(zz=0;zz<j;zz++){   if(c[k][z-zz]!=c[id][i+zz])break;}      if(zz==j)   {      ok=1;  break;   }  }   if(!ok)   //有一个不行 这个子串就不行了   break;  }if(k==n){//坑了好久 枚举的起点放在长度外面了 造成被更新了...日了狗了 re=max(re,j);  break; } }}        printf("%d\n",re); }return 0;}*/

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