UVA 10829 L-Gap Substrings（后缀数组）

最新推荐文章于 2020-11-27 20:26:56 发布

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最新推荐文章于 2020-11-27 20:26:56 发布

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分类专栏：后缀数组文章标签：后缀数组 LCP

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本文探讨了如何通过枚举U的长度并采用类POJ3693的方法，计算给定字符串中UVU模式出现的次数。通过构建SA数组、获取高度数组以及初始化RMQ数组，实现O(nlogn)的复杂度进行高效匹配。

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题意就是找出一个串中形如UVU格式的字符串有几个。

枚举U的长度L，利用类似poj3693一样的做法，对于位置i和位置i+g+L先向后匹配再向前匹配，为了不重复，像后匹配长度不能超过L，向前匹配不超过L-1，然后答案加上匹配总长度-L+1。

总复杂度o（nlogn）

AC代码：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll long long
#define ull unsigned long long
#define eps 1e-8
#define NMAX 100100
#define MOD 1000000007
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
template<class T>
inline void scan_d(T &ret)
{
    char c;
    int flag = 0;
    ret=0;
    while(((c=getchar())<'0'||c>'9')&&c!='-');
    if(c == '-')
    {
        flag = 1;
        c = getchar();
    }
    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
    if(flag) ret = -ret;
}

char ch[NMAX];
int s[NMAX],t[NMAX],t2[NMAX],c[NMAX],sa[NMAX];
int rank[NMAX],height[NMAX],dp[NMAX][20];
void build_sa(int n, int m)
{
    int i, *x = t, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = s[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i-1];
    for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1)
    {
        int p = 0;
        for(i = n-k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[i]]++;
        for(i = 1; i < m; i++) c[i] += c[i-1];
        for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;x[sa[0]] = 0;
        for(i = 1; i < n; i++)
            x[sa[i]] = y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
        if(p >= n) break;
        m = p;
    }
}

void getHeight(int n)
{
    int i,k = 0;
    for(i = 0; i < n; i++) rank[sa[i]] = i;
    for(i = 0; i < n; i++)
    {
        if(k) k--;
        if(rank[i] == 0) continue;
        int j = sa[rank[i]-1];
        while(s[i+k] == s[j+k]) k++;
        height[rank[i]] = k;
    }
}

void init_RMQ(int n)
{
    n--;
    for(int i = 1; i <= n; i++) dp[i][0] = height[i];
    for(int l = 1; (1<<l) <= n; l++)
        for(int i = 1; i + (1<<l) - 1 <= n; i++)
            dp[i][l] = min(dp[i][l-1],dp[i+(1<<(l-1))][l-1]);
}

int RMQ(int l, int r)
{
    if(r < l) swap(l,r);
    l++;
    int k = (int)log2(r-l+1.0);
    return min(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
#ifdef GLQ
    freopen("input.txt","r",stdin);
//    freopen("o1.txt","w",stdout);
#endif // GLQ
    int cas = 1,g,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%s",&g,ch);
        printf("Case %d: ",cas++);
        int nct = 0,len = strlen(ch),ha = 26;
        for(int i = 0 ; i < len; i++)
        {
            s[nct++] = ch[i]-'a'+1;
            ha = min(ha,ch[i]-'a');
        }
        s[nct++] = 27;
        for(int i = len-1; i >= 0; i--)
            s[nct++] = ch[i]-'a'+1;
        s[nct++] = 28;
        build_sa(nct,29);
        getHeight(nct);
        init_RMQ(nct);
        ll ans = 0;
        for(int l = 1; 2*l + g <= len; l++)
            for(int i = 0; i + g + l < len; i+=l)
            {
                int j = i+g+l;
                if(ch[i] != ch[j]) continue;
                ll tmp = min(l,RMQ(rank[i],rank[j])) + min(l-1,RMQ(rank[nct-i-1],rank[nct-j-1]));
                if(tmp < l) continue;
                ans += tmp-l+1;
            }
        printf("%lld\n",ans);
    }
    return 0;
}