【bzoj1717】Milk Patterns 后缀数组 + （二分||单调队列）

来源：互联网发布：数据分析师学历编辑：程序博客网时间：2024/05/17 06:54

这题看完之后想用单调队列……结果调了一个下午……
DQS学长看了几秒钟之后说是可以二分……
然后……
OrzOrzOrz

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>using namespace std;const int MAXN = 100000 + 5;int sa[MAXN],rank[MAXN],tmp[MAXN];int n,m,k;int c[MAXN];bool cmp_sa(int a,int b){    if(rank[a] != rank[b])        return rank[a] < rank[b];    else    {        int x = a + k <= n ? rank[a + k] : -1;        int y = b + k <= n ? rank[b + k] : -1;        return x < y;    }}void make_sa(int s[]){    for(int i = 0;i <= n;i ++)    {        rank[i] = i < n ? s[i] : -1;        sa[i] = i;    }    for(k = 1;k <= n;k <<= 1)    {        sort(sa,sa + n + 1,cmp_sa);        tmp[sa[0]] = 0;        for(int i = 1;i <= n;i ++)            tmp[sa[i]] = tmp[sa[i - 1]] + cmp_sa(sa[i - 1],sa[i]);        for(int i = 0;i <= n;i ++)            rank[i] = tmp[i];    }    return;}int lcp[MAXN];void make_lcp(int s[]){    for(int i = 1;i < n;i ++)        rank[sa[i]] = i;    int h = 0;    lcp[0] = 0;    for(int i = 0;i < n;i ++)    {        int j = sa[rank[i] - 1];        if(h)            h--;        for(;i + h < n,j + h < n;h ++)            if(s[i + h] != s[j + h])                break;        lcp[rank[i] - 1] = h;    }    return;}bool can(int x){    int ans = 0,num = 0;    for(int i = 1;i <= n;i ++)    {        num = (num + 1) * (lcp[i] >= x);        ans = max(num,ans);    }    return ans >= m - 1;}int div(){    int ans;    int l = 1,r = n;    while(l <= r)    {        int mid = (l + r) >> 1;        if(can(mid))            ans = mid,l = mid + 1;        else            r = mid - 1;    }    return ans;}int main(){    scanf("%d %d",&n,&m);    for(int i = 0;i < n;i ++)        scanf("%d",&c[i]);    make_sa(c);    make_lcp(c);    printf("%d\n",div());    return 0;}

D(y)Q(d)S(c)：“一般后缀数组或者高度数组都是套个二分什么的，把lcp数组切成一块一块的之后就可以简单的验证了……”
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