UOJ #219 BZOJ 4650 luogu P1117 [NOI2016]优秀的拆分 (后缀数组、ST表)

连NOI Day1T1都不会做。。。看了题解都写不出来还要抄Claris的代码。。

题目链接: (luogu)https://www.luogu.org/problemnew/show/P1117

(bzoj)https://www.lydsy.com/JudgeOnline/problem.php?id=4650

(uoj)http://uoj.ac/problem/219

题解:

\(f[i]\)表示以\(i\)结束的\(AA\)型子串个数，\(g[i]\)表示以\(i\)开始的\(AA\)型子串个数

怎么求\(f,g\)?

打破思维定势，谁说必须要一个一个求呢

分长度来求

枚举长度\(L\), 处理所有长度为\(2L\)的\(AA\)型子串对\(f\)和\(g\)的贡献

如果每隔\(L\)的长度放一个打点计时器，呸，关键点

那么任何\(AA\)型子串都会经过两个相邻关键点

首先肯定要满足这两个关键位置上的字符一样

在这个基础上求出往前往后最多多少个一样的

这个就转化成了LCP和LCS问题，并且两个相邻关键点对\(f,g\)数组的影响是区间+1，使用差分前缀和解决

然后推一推就行了，注意+1-1不要推错

时间复杂度为调和级数，\(O(n\log n)\)

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define llong long long
using namespace std;

const int N = 1<<15;
const int LGN = 15;
const int S = 26;
int log2[N+3];
struct SparseTable
{
    int n;
    int str[N+3];
    int rk[N+3];
    int tmp[N+3];
    int height[N+3];
    int h[N+3];
    int sa[N+3];
    int wb[N+3];
    int mini[N+3][LGN+3];
    void get_sa()
    {
        int *x = rk,*y = tmp;
        for(int i=0; i<=S; i++) wb[i] = 0;
        for(int i=1; i<=n; i++) wb[x[i]=str[i]]++;
        for(int i=1; i<=S; i++) wb[i] += wb[i-1];
        for(int i=n; i>=1; i--) sa[wb[x[i]]--] = i;
        int s = S,p = 0;
        for(int j=1; p<n; j<<=1)
        {
            p = 0;
            for(int i=n-j+1; i<=n; i++) y[++p] = i;
            for(int i=1; i<=n; i++) if(sa[i]>j) y[++p] = sa[i]-j;
            for(int i=1; i<=s; i++) wb[i] = 0;
            for(int i=1; i<=n; i++) wb[x[y[i]]]++;
            for(int i=1; i<=s; i++) wb[i] += wb[i-1];
            for(int i=n; i>=1; i--) sa[wb[x[y[i]]]--] = y[i];
            swap(x,y);
            p = 1; x[sa[1]] = 1;
            for(int i=2; i<=n; i++) x[sa[i]] = (y[sa[i]]==y[sa[i-1]] && y[sa[i]+j]==y[sa[i-1]+j]) ? p : ++p;
            s = p;
        }
        for(int i=1; i<=n; i++) rk[sa[i]] = i;
        for(int i=1; i<=n; i++)
        {
            h[i] = h[i-1]==0 ? 0 : h[i-1]-1;
            while(i+h[i]<=n && sa[rk[i-1]]+h[i]<=n && str[i+h[i]]==str[sa[rk[i]-1]+h[i]])
            {
                h[i]++;
            }
        }
        for(int i=1; i<=n; i++) height[i] = h[sa[i]];
        for(int i=1; i<=n; i++) mini[i][0] = height[i];
        for(int j=1; j<=LGN; j++)
        {
            for(int i=1; i+(1<<j)-1<=n; i++)
            {
                mini[i][j] = min(mini[i][j-1],mini[i+(1<<j-1)][j-1]);
            }
        }
    }
    int querymin(int lb,int rb)
    {
        int g = log2[rb-lb+1];
        return min(mini[lb][g],mini[rb-(1<<g)+1][g]);
    }
    int LCP(int x,int y)
    {
        if(x==y) return n-x+1;
        if(rk[x]>rk[y]) swap(x,y);
        return querymin(rk[x]+1,rk[y]);
    }
    void clear()
    {
        for(int i=1; i<=n; i++) str[i] = rk[i] = tmp[i] = height[i] = h[i] = sa[i] = wb[i] = 0;
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<=LGN; j++)
            {
                mini[i][j] = 0;
            }
        }
    }
} s1,s2;
llong f[N+3];
llong g[N+3];
char a[N+3];
int n;

int LCP(int x,int y) {return s1.LCP(x,y);}
int LCS(int x,int y) {return s2.LCP(n+1-x,n+1-y);}

void preprocess()
{
    log2[1] = 0; for(int i=2; i<=N; i++) log2[i] = log2[i>>1]+1;
}

void clear()
{
    s1.clear(); s2.clear();
    for(int i=0; i<=n+1; i++) a[i] = 0,f[i] = g[i] = 0ll;
}

int main()
{
    preprocess();
    int T; scanf("%d",&T);
    while(T--)
    {
        scanf("%s",a+1); n = strlen(a+1); for(int i=1; i<=n; i++) a[i]-=96;
        for(int i=1; i<=n; i++) s1.str[i] = a[i]; s1.n = n;
        s1.get_sa();
        for(int i=1; i<=n; i++) s2.str[i] = a[n+1-i]; s2.n = n;
        s2.get_sa();
        for(int i=1; i+i<=n; i++)
        {
            for(int j=i+i; j<=n; j+=i)
            {
                if(a[j]==a[j-i])
                {
                    int lb = j-LCS(j,j-i)+1,rb = j+LCP(j,j-i)-1;
                    lb = max(lb+i-1,j); rb = min(rb,j+i-1);
                    if(lb<=rb)
                    {
                        f[lb]++; f[rb+1]--;
                        g[lb-i-i+1]++; g[rb+1-i-i+1]--;
                    }
                }
            }
        }
        for(int i=1; i<=n; i++) f[i] += f[i-1],g[i] += g[i-1];
        llong ans = 0ll;
        for(int i=1; i<n; i++)
        {
            llong tmp = f[i]*g[i+1];
            ans += tmp;
        }
        printf("%lld\n",ans);
        clear();
    }
    return 0;
}

标签：AA,子串,P1117,int,luogu,clear,219,sa,include
来源： https://www.cnblogs.com/suncongbo/p/11032784.html