POJ

来源：互联网发布：淘宝大学的证书编辑：程序博客网时间：2024/05/21 10:56

题意

给出长度为 n 的正整数串，求重复出现了至少 K 次的串的最大长度。

思路

后缀数组模板题。

链接

https://cn.vjudge.net/contest/177933#problem/G

代码

#include<cstdio>#include<iostream>using namespace std;const int maxn = 20010;int s[maxn];int sa[maxn];int t1[maxn], t2[maxn], c[maxn];int rank[maxn], height[maxn];void build_sa(int s[], int n, int m){    int *x = t1, *y = t2;    for(int i = 0; i < m; i++) c[i] = 0;    for(int i = 0; i < n; i++) c[x[i] = s[i]] ++;    for(int i = 1; i < m; i++) c[i] += c[i-1];    for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;    for(int j = 1; j <= n; j <<= 1)    {        int p = 0;        for(int i = n-j; i < n; i++) y[p++] = i;        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;        for(int i = 0; i < m; i++) c[i] = 0;        for(int i = 0; i < n; i++) c[x[y[i]]]++;        for(int i = 1; i < m; i++) c[i] += c[i-1];        for(int i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];        swap(x, y);        p = 1, x[sa[0]] = 0;        for(int i = 1; i < n; i++)            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;        if(p >= n) break;        m = p;    }}void getHeight(int s[], int n){    int k = 0;    for(int i = 0; i <= n; i++) rank[sa[i]] = i;    for(int i = 0; i < n; i++)    {        if(k) k--;        int j = sa[rank[i]-1];        while(s[i+k] == s[j+k]) k++;        height[rank[i]] = k;    }}bool check(int n, int k, int t){    int num = 1;    for(int i = 2; i <= n; i++)    {        if(height[i] >= t)        {            num ++;            if(num >= k) return true;        }        else num = 1;    }    return false;}int main(){    int n, k;    while(scanf("%d %d", &n, &k) == 2)    {        int Max = 0;        for(int i = 0; i < n; i++)        {            scanf("%d", &s[i]);            Max = max(Max, s[i]);        }        s[n] = 0;        build_sa(s, n+1, Max+1);        getHeight(s, n);        int l = 0, r = n;        int ans = 0;        while(l <= r)        {            int mid = (l + r) >> 1;            if(check(n, k, mid))            {                ans = mid;                l = mid + 1;            }            else r = mid - 1;        }        printf("%d\n", ans);    }    return 0;}

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