SPOJ220 Relevant Phrases of Annihilation

http://www.spoj.com/problems/PHRASES/

题意:给n个串，求n个串里面都有2个不重叠的最长的字串长度。

思路:二分答案，然后就可以嘿嘿嘿

PS:辣鸡题目毁我青春，一开始二分的时候ans没有赋初值为0，结果没答案的时候就会输出奇怪的数字T_T,其实主要还是怪我不小心。。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 int len,num[],ws[],wv[],wa[],wb[],h[],sa[],rank[];
 int f[][],N,b[];
 char s[];
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-')f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';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 *t,*x=wa,*y=wb,i,j,k,p;
     for (i=;i<m;i++) ws[i]=;
     for (i=;i<n;i++) x[i]=r[i];
     for (i=;i<n;i++) ws[x[i]]++;
     for (i=;i<m;i++) ws[i]+=ws[i-];
     for (i=n-;i>=;i--) sa[--ws[x[i]]]=i;
     for (j=,p=;p<n;j*=,m=p){
         for (p=,i=n-j;i<n;i++) y[p++]=i;
         for (i=;i<n;i++) if (sa[i]>=j) y[p++]=sa[i]-j;
         for (i=;i<m;i++) ws[i]=;
         for (i=;i<n;i++) wv[i]=x[y[i]];
         for (i=;i<n;i++) ws[wv[i]]++;
         for (i=;i<m;i++) ws[i]+=ws[i-];
         for (i=n-;i>=;i--) sa[--ws[wv[i]]]=y[i];
         for (t=x,x=y,y=t,i=,x[sa[]]=,p=;i<n;i++){
             x[sa[i]]=cmp(y,sa[i-],sa[i],j)?p-:p++;
         }
     }
 }
 void cal(int *r,int n){
     int i,j,k=;
     for (int i=;i<=n;i++) rank[sa[i]]=i;
     for (int i=;i<n;h[rank[i++]]=k){
         for (k?k--:,j=sa[rank[i]-];r[j+k]==r[i+k];k++);
     }
 }
 bool check(int mid){
     int l,r;
     for (int i=;i<=len;i++){
         l=i;
         while (l<=len&&h[l]<mid) l++;
         r=l;
         while (r<=len&&h[r]>=mid) r++;
         for (int j=;j<=N;j++) f[j][]=0x3f3f3f3f,f[j][]=-0x3f3f3f3f;
         for (int j=l-;j<=r-;j++){
             int k=b[sa[j]];
             if (k==N+) continue;
             f[k][]=std::min(f[k][],sa[j]);
             f[k][]=std::max(f[k][],sa[j]);
         }
         int j=;
         for (j=;j<=N;j++){
             if (f[j][]==0x3f3f3f3f) break;
             if (f[j][]-f[j][]<mid) break;
         }
         if (j>N) return ;
         i=r;
     }
     return ;
 }
 void solve(){
     int l=,r=,ans=;
     while (l<=r){
         int mid=(l+r)>>;
         if (check(mid)) ans=mid,l=mid+;
         else r=mid-;
     }
     printf("%d ",ans);
 }
 int main(){
     int T=read();
     while (T--){
         len=;
         int n=read();
         for (int i=;i<=n;i++){
             scanf("%s",s+);
             int Len=strlen(s+);
             for (int j=;j<=Len;j++)
              num[len]=s[j],b[len++]=i;
             if (i<n) num[len]=+i,b[len++]=n+;
         }
         num[len]=;b[len]=n+;
         da(num,sa,len+,);
         cal(num,len);
         N=n;
         solve();
     }
 }