xmu 1337 后缀数组 + 暴力匹配

http://acm.xmu.edu.cn/JudgeOnline/problem.php?id=1337

题意：在一个串中求出一个长度为n的、重复了m次的子串（不能重叠）。

思路：一直以为正解是后缀数组加其它高效的搜索方法。一水才知道，就是纯后缀数组+暴力匹配。

先用后缀数组求出 sa,height两个数组的值，当height[i]=n时，取出当前子串，进行暴力匹配。囧~~~

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int maxn = 101000;
int wn[maxn],wa[maxn],wb[maxn],wv[maxn],a[maxn],sa[maxn],rank[maxn],height[maxn];
char r[maxn];

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);
}
int main()
{
    int i,j,k,n,m,cnt;
    char as[2005],*p;
    while(scanf("%d %d",&n,&m)==2)
    {
        scanf("%s",r);
        for(i = 0; r[i]; ++ i) a[i] = static_cast<int>(r[i]);
        a[i] = 0;
        da(a,sa,i+1,256);
        calheight(a,sa,i);
        for(j = 1; j <= i; ++ j)if(height[j]==n)
        {
            for(k = 0; k < n; ++ k) as[k] = r[sa[j]+k];
            as[k] = 0;
            p = r;
            p = strstr(p,as);
            p += n;
            cnt = 1;
            while(cnt<m){ p = strstr(p,as); if(p==NULL)break; cnt++; p += n; }
            if(cnt>=m)break;
        }
        puts(as);
    }return 0;
}