POJ-3261 Milk Patterns (后缀数组 最长k次重复子串)

题目：

http://poj.org/problem?id=3261

题意：

求n天牛奶质量序列中的最长k次重复子串，子串之间可重复。

思路：

标准后缀数组的模版题，利用后缀数组的height数组可以判断排名连续的k个后缀子串的前缀相同，则满足条件。

因为序列最多有2000个，所以只要对于整个序列离散化之后求后缀数组。然后二分枚举子串长度res，通过height数组扫一遍判断是否存在大于等于k个后缀子串的前缀有大于res个长度相同即可。

代码：

#include <bits/stdc++.h>#define N 20005using namespace std;int n,m,sum,res,flag,k;int seq[N], sa[N], ranks[N], height[N];int wwa[N], wwb[N], wws[N], wwv[N];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 i, j, p, *x = wwa, *y = wwb;    for (i = 0; i < m; ++i) wws[i] = 0;    for (i = 0; i < n; ++i) wws[x[i]=r[i]]++;    for (i = 1; i < m; ++i) wws[i] += wws[i-1];    for (i = n-1; i >= 0; --i) sa[--wws[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 < n; ++i) wwv[i] = x[y[i]];        for (i = 0; i < m; ++i) wws[i] = 0;        for (i = 0; i < n; ++i) wws[wwv[i]]++;        for (i = 1; i < m; ++i) wws[i] += wws[i-1];        for (i = n-1; i >= 0; --i) sa[--wws[wwv[i]]] = y[i];        for (swap(x, y), p = 1, x[sa[0]] = 0, 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) ranks[sa[i]] = i;    for (i = 0; i < n; height[ranks[i++]] = k)        for (k?k--:0, j = sa[ranks[i]-1]; r[i+k] == r[j+k]; k++);}void solve(){    //初始序0~n，最大值小于m。    seq[n]=0;    da(seq,sa,n+1,m);    calheight(seq,sa,n);}int temp[N];map<int,int>mp;int main(){    int i,j,kk,cas,T,t,x,y,z;    while(scanf("%d%d",&n,&k)!=EOF)    {        mp.clear();        for(i=0;i<n;i++)        {            scanf("%d",&seq[i]);            temp[i]=seq[i];        }        sort(temp,temp+n);        m=unique(temp,temp+n)-temp;        for(i=0;i<m;i++)            mp[temp[i]]=i+1;        for(i=0;i<n;i++)            seq[i]=mp[seq[i]];        solve();        x=1;y=n;        while(x<=y)        {            z=(x+y)/2;            flag=false;            for(i=2;i<=n&&!flag;i++)            {                t=1;                while(height[i]>=z)i++,t++;                if(t>=k)flag=true;            }            if(flag)                x=z+1,res=z;            else                y=z-1;        }        printf("%d\n",res);    }    return 0;}