hdu 2328 Corporate Identity

http://acm.hdu.edu.cn/showproblem.php?pid=2328 

题意：求多个字符串的最长公共子串，若有多个输出字典序最小。

思路：先进行后缀数组操作。然后二分子串长度，根据height和sa数组的值，判断当前长度是否合适。

当然，这里面还有hash映射的。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int maxn = 910000;
int wn[maxn],wa[maxn],wb[maxn],wv[maxn],as[maxn],sa[maxn],rank[maxn],height[maxn];
int sta[maxn],hash[5000],bs[maxn],n,ansx;
char r[maxn],ans[1000];
int Max(int x,int y){return x>y?x:y;}
int Min(int x,int y){return x>y?y:x;}
int 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 i,j,p,*x=wa,*y=wb,*t;
    for(i = 0; i < m; ++ i) wn[i] = 0;
    for(i = 0; i < n; ++ i) wn[x[i]=r[i]]++;
    for(i = 1; i < m; ++ i) wn[i] += wn[i-1];
    for(i = n - 1; i >= 0; --i) sa[--wn[x[i]]] = i;
    for(p = 1,j = 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) wn[i] = 0;
        for(i = 0; i < n; ++ i) wn[wv[i]=x[y[i]]]++;
        for(i = 1; i < m; ++ i) wn[i] += wn[i-1];
        for(i = n - 1; i >= 0; --i) sa[--wn[wv[i]]] = y[i];
        for(t = x,x = y,y = t,x[sa[0]] = 0,p=1,i=1; i<n; ++i)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
}
void calheight(int *r,int *sa,int n)
{
    int i,j,k = 0;
    for(i = 1; i <= n; ++ i){ rank[sa[i]] = i; height[i] = 0;}
    for(i = 0; i < n; height[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)
{
    int i,j=0,k = 0;
    memset(hash,-1,sizeof(hash));
    hash[0] = 0;--L;
    for(i = 1; i < L; ++ i)
    {
        if(height[i]>=x){
             if(!j){ k+= hash[bs[sa[i-1]]]; hash[bs[sa[i-1]]] = 0; j = 1; }
             k += hash[bs[sa[i]]]; hash[bs[sa[i]]] = 0;
        }else{
            k *= -1;
            if(k>=n){
                if(x>ansx){
                    for(j = 0; j < x; ++ j) 
                        ans[j] = r[sa[i-1]+j];
                    ans[j] = 0;
                    ansx = x;
                }
                return 1;
            }
            k = 0;
            memset(hash,-1,sizeof(hash));
            hash[0] = 0;
            j = 0;
        }
    }return 0;
}
int main()
{
    int i,j,k,m,cnt,L,t,mixn;
    char str[1000];
    while(scanf("%d",&n)!=EOF&&n)
    {
        k = 199;
        L = 0; mixn = 200;
        for(j = 0; j < n; ++ j)
        {
            sta[j] = L;
            scanf("%s",str);
            strcpy(r+L,str);
            for(i = 0; str[i]; ++ i) as[L++] = static_cast<int>(str[i]);
            if(i<mixn)mixn = i;
            r[L] = '#'; as[L++] = ++k;
        }
        sta[j] = L;
        r[L-1] = 0;
        as[L-1] = 0;
        da(as,sa,L,k);
        calheight(as,sa,L-1);
        for(i=j=0; i < L; ++ i) if(i<sta[j]) bs[i] = j; else bs[i] = ++j;
        for(i = 1; i <= n; ++ i) bs[sta[i]-1] = 0;
        int low = 0,high = mixn+1,mid;
        ansx = 0;
        while(low<high-1)
        {
            mid = (low + high)>>1;
            if(check(mid,L))low = mid;
            else high = mid;            
        }
        if(ansx) puts(ans);
        else puts("IDENTITY LOST");
    }return 0;
}