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[500005],ws[500005],wv[500005],wa[500005],wb[500005],h[500005],sa[500005],rank[500005];
int f[500005][2],N,b[500005];
char s[500005];
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 *t,*x=wa,*y=wb,i,j,k,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 (j=1,p=1;p<n;j*=2,m=p){
        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,x[sa[0]]=0,p=1;i<n;i++){
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
        }
    }
}
void cal(int *r,int n){
    int i,j,k=0;
    for (int i=1;i<=n;i++) rank[sa[i]]=i;
    for (int i=0;i<n;h[rank[i++]]=k){
        for (k?k--:0,j=sa[rank[i]-1];r[j+k]==r[i+k];k++);
    }
}
bool check(int mid){
    int l,r;
    for (int i=1;i<=len;i++){
        l=i;
        while (l<=len&&h[l]<mid) l++;
        r=l;
        while (r<=len&&h[r]>=mid) r++;
        for (int j=1;j<=N;j++) f[j][0]=0x3f3f3f3f,f[j][1]=-0x3f3f3f3f;
        for (int j=l-1;j<=r-1;j++){
            int k=b[sa[j]];
            if (k==N+1) continue;
            f[k][0]=std::min(f[k][0],sa[j]);
            f[k][1]=std::max(f[k][1],sa[j]);
        }
        int j=0;
        for (j=1;j<=N;j++){
            if (f[j][0]==0x3f3f3f3f) break;
            if (f[j][1]-f[j][0]<mid) break;
        }
        if (j>N) return 1;
        i=r;
    }
    return 0;
}
void solve(){
    int l=1,r=20000,ans=0;
    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();
    while (T--){
        len=0;
        int n=read();
        for (int i=1;i<=n;i++){
            scanf("%s",s+1);
            int Len=strlen(s+1);
            for (int j=1;j<=Len;j++)
             num[len]=s[j],b[len++]=i;
            if (i<n) num[len]=290+i,b[len++]=n+1; 
        }
        num[len]=0;b[len]=n+1;
        da(num,sa,len+1,310);
        cal(num,len);
        N=n;
        solve();
    }
}