POJ1226

题目

Substrings
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 14254 Accepted: 5064
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 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

2
3
ABCD
BCDFF
BRCD
2
rose
orchid
Sample Output

2
2

题目大意

给一坨字符串，求一个最长的串的长度，使得这个串在每一个串或这个串倒过来之后是其子串（连续的）

做法

首先把每一个串以及其倒过来之后的串连接在一起，然后用一个数组来记录每一个位置属于哪一个字符串，然后二分求可行性就可以了

贴代码

#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#define fo(i,a,b) for(i=a;i<=b;i++)using namespace std;const int ma=25005;int sa[ma],height[ma],rank[ma],x[ma],y[ma],buc[ma],bi[ma],go[ma];char s[ma],s1[ma];bool bz[ma];int i,j,k,l,r,n,t,len,now,m,p,o,t1,t2,ans,zo,mi,mid;void gesa(){    m=96+26;    memset(y,0,sizeof(y));    fo(i,0,len-1) x[i]=s[i];    fo(i,1,m) buc[i]=0;    fo(i,0,len-1) buc[x[i]]++;    fo(i,1,m) buc[i]+=buc[i-1];    fo(i,0,len-1) sa[buc[x[i]]--]=i;    for(k=1;k<len;k=k*2){        p=0;        fo(i,len-k,len-1) y[++p]=i;        fo(i,1,len) if (sa[i]>=k) y[++p]=sa[i]-k;        fo(i,1,m) buc[i]=0;        fo(i,0,len-1) buc[x[i]]++;        fo(i,1,m) buc[i]+=buc[i-1];        for(i=len;i>=1;i--) sa[buc[x[y[i]]]--]=y[i];        fo(i,0,len-1) y[i]=x[i];        x[sa[1]]=1;        fo(i,2,len){            if (y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k]) x[sa[i]]=x[sa[i-1]];            else x[sa[i]]=x[sa[i-1]]+1;        }        m=x[sa[len]];        if (m>=len) break;    }    fo(i,1,len) rank[sa[i]]=i;}void gehei(){    k=0;    fo(i,0,len-1){        if (rank[i]==1){            height[rank[i]]=0;            continue;        }        if (k) k--;        t1=i+k; t2=sa[rank[i]-1]+k;        while (t1<len && t2<len){            if (s[t1]=='#' || s[t2]=='#') break;            if (s[t1]!=s[t2]) break;            k++; t1++; t2++;        }        height[rank[i]]=k;    }}bool geans(int k){    now=0;    memset(bz,false,sizeof(bz));    fo(i,1,len)    if (s[sa[i]]!='#')    {        if (height[i]<k){            fo(j,1,now) bz[go[j]]=false;            now=1;            go[1]=bi[sa[i]];            bz[bi[sa[i]]]=true;            continue;        }        if (bz[bi[sa[i]]]==false){            go[++now]=bi[sa[i]];            bz[bi[sa[i]]]=true;        }        if (now==zo) return true;    }    return false;}void getwo(){    l=0;    r=mi;    while (l<r){        mid=(l+r)/2;        if (geans(mid)) l=mid+1; else r=mid;    }    if (geans(l)==false) l--;    printf("%d\n",l);}int main(){//  freopen("1226.in","r",stdin);//  freopen("1226.out","w",stdout);    scanf("%d",&t);    fo(o,1,t){        scanf("%d",&n);        now=0;        zo=0;        mi=ma;        fo(i,1,n){            scanf("%s",&s1);            len=strlen(s1);            mi=min(len,mi);            zo++;            fo(j,0,len-1)            {                s[now]=s1[j];                bi[now++]=zo;            }            s[now++]='#';            for(j=len-1;j>=0;j--)            {                s[now]=s1[j];                bi[now++]=zo;            }            s[now++]='#';        }        len=now;        gesa();        gehei();        getwo();    }    return 0;}