[poj][3261][Milk Patterns]

题目：http://poj.org/problem?id=3261

第一道后缀数组。二分答案，然后遍历height数组，判断该答案是否出现次数大于k次。

View Code

#include <cstdio>

#include <cstring>

#include <map>



using namespace std;



const int N = 1000000+10;

int ua[N], ub[N], us[N], n, m, arr[N], sa[N], t, cnt;

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 n,int m){

    int i,j,p,*x=ua,*y=ub,*t;

    for (i=0;i<m;i++) us[i]=0;

    for (i=0;i<n;i++) us[x[i]=r[i]]++;

    for (i=1;i<m;i++) us[i]+=us[i-1];

    for (i=n-1;i>=0;i--) sa[--us[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++) us[i]=0;

        for (i=0;i<n;i++) us[x[i]]++;

        for (i=1;i<m;i++) us[i]+=us[i-1];

        for (i=n-1;i>=0;i--) sa[--us[x[y[i]]]]=y[i];

        for (t=x,x=y,y=t,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++;

    }

}

int rank[N],height[N];

void calheight(int *r,int n){

    int i,j,k=0;

    for (i=1;i<=n;i++)rank[sa[i]]=i;

    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++);

}

void solve(int k){

    int i, tot, maxx=0;

    int low=1, high=n, mid;

    while (low <= high){

        mid = low + (high - low) / 2;

        for (tot=1, i=2; i<=n; i++){

            if (height[i]>=mid)tot++;

            else tot = 1;

            if (tot >= k) break;

        }

        if (tot >= k)low=mid+1,maxx=mid;

        else high=mid-1;

    }

    printf("%d\n", maxx);

}



int main(){

    //freopen("D:/a.txt", "r", stdin);

    while (~scanf("%d%d", &n, &m)){

        memset(arr, 0, sizeof(arr));

        map<int,int>ma;

        cnt = 1;

        for (int i=0; i<n; i++){

            scanf("%d", &t);

            if (ma.find(t)==ma.end())ma[t]=cnt++;

            arr[i] = ma[t];

        }

        arr[n] = 0;

        da(arr, n+1, cnt); //算后缀数组的时候要把最后一个算在内，排名才是从1开始的，因为空串排在第0个。

        calheight(arr, n);

        solve(m);

    }

    return 0;

}