HDU 1238 Substrings(求公共正反向连续子串)

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8504 Accepted Submission(s): 3931



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
There should be one line per test case containing the length of the largest string found.

[align=left]Sample Input[/align]

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

[align=left]Sample Output[/align]

2
2

[align=left]Author[/align]
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

;
int len
;
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

;
int len
;
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;
}
*/