POJ 1225 Substrings

http://poj.org/problem?id=1226

题意：给定n个串。求一个最长的串，使得这个串或者其反串在每个串中都出现过？

思路：先在大串里面加入正反串，然后二分，判定即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
int num[200005],ws[200005],wv[200005],wb[200005],wa[200005];
int n,m,rank[200005],sa[200005],h[200005],b[200005];
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
bool cmp(int *r,int a,int b,int l){
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int *r,int *sa,int n,int m){
    int *y=wb,*x=wa,*t,i,j,p;
    for (i=0;i<m;i++) ws[i]=0;
    for (i=0;i<n;i++) x[i]=r[i];
    for (i=0;i<n;i++) ws[x[i]]++;
    for (i=1;i<m;i++) ws[i]+=ws[i-1];
    for (i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;
    for (p=1,j=1;p<n;m=p,j*=2){
        for (p=0,i=n-j;i<n;i++) y[p++]=i;
        for (i=0;i<n;i++) if (sa[i]>=j) y[p++]=sa[i]-j;
        for (i=0;i<m;i++) ws[i]=0;
        for (i=0;i<n;i++) wv[i]=x[y[i]];
        for (i=0;i<n;i++) ws[wv[i]]++;
        for (i=1;i<m;i++) ws[i]+=ws[i-1];
        for (i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];
        for (t=x,x=y,y=t,i=1,p=1,x[sa[0]]=0;i<n;i++)
         x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;
    }
}
void cal(int *r,int n){
    int i,j,k=0;
    for (i=1;i<=n;i++) rank[sa[i]]=i;
    for (i=0;i<n;h[rank[i++]]=k)
     for (k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
}
bool check(int x){
    int L,R;
    int hash[105];
    for (int i=1;i<=n;i++){
        L=i;
        while (L<=n&&h[L]<x) L++;
        if (L>n) break;
        R=L;
        while (R<=n&&h[R]>=x) R++;
        memset(hash,0,sizeof hash);
        for (int j=L-1;j<=R-1;j++)
         if (b[sa[j]]<=m)
          hash[b[sa[j]]]=1;
        int j=0;
        for (j=1;j<=m;j++)
         if (!hash[j]) break;
        if (j>m) return 1; 
        i=R; 
    }
    return 0;
}
void solve(){
    int l=0,r=100,ans;
    while (l<=r){
        int mid=(l+r)>>1;
        if (check(mid)) ans=mid,l=mid+1;
        else r=mid-1;
    }
    printf("%d
",ans);
}
int main(){
    int T=read();
    char s[200005];
    while (T--){
        m=read();n=0;int sx=130;
        for (int i=1;i<=m;i++){
            scanf("%s",s);
            int len=strlen(s);
            for (int j=0;j<len;j++)
             num[n]=s[j],b[n++]=i;
            num[n]=sx++;b[n++]=n+1; 
            for (int j=len-1;j>=0;j--)
             num[n]=s[j],b[n++]=i;
            num[n]=sx++;b[n++]=n+1; 
        }
        num[n]=0;b[n]=n+1;
        da(num,sa,n+1,sx+10);
        cal(num,n);
        solve();
    }
}