• POJ 3261 Milk Patterns 后缀数组


    用后缀数组求重复出现至少k次的可重叠最长子串的长度, 当然是可以用hash搞的,用后缀数组的话,只要在分组之后看看个数是不是大于等于k

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <climits>
    #include <string>
    
    using namespace std;
     
    #define MP make_pair
    #define PB push_back
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    typedef pair<double, double> PDD;
    const int INF = INT_MAX / 3;
    const double eps = 1e-8;
    const LL LINF = 1e17;
    const double DINF = 1e60;
    const int maxn = 1e6 + 10;
    
    //以下是倍增法求后缀数组
    int wa[maxn], wb[maxn], wv[maxn], ws[maxn];
    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 *sa, int n, int m) {
        int i, j, p, *x = wa, *y = wb, *t;
        for(i = 0; i < m; i++) ws[i] = 0;
        for(i = 0; i < n; i++) ws[x[i] = r[i]]++;
        for(i = 1; i < m; i++) ws[i] += ws[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
        for(j = 1, p = 1; p < n; j <<= 1, 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 < n; i++) wv[i] = x[y[i]];
            for(i = 0; i < m; i++) ws[i] = 0;
            for(i = 0; i < n; i++) ws[wv[i]]++;
            for(i = 0; i < m; i++) ws[i] += ws[i - 1];
            for(i = n - 1; i >= 0; i--) sa[--ws[wv[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++;
        }
    }
    
    //以下是求解height 数组
    int height[maxn], Rank[maxn];
    void calheight(int *r, int *sa, 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++) ;
    }
    
    int n, k, val[maxn], sa[maxn];
    
    bool ok(int len) {
        int nowcnt = 1;
        for(int i = 1; i <= n; i++) {
            if(height[i] >= len) {
                nowcnt++;
                if(nowcnt >= k) return true;
            }
            else nowcnt = 1;
        }
        return false;
    }
    
    int main() {
        while(scanf("%d%d", &n, &k) != EOF) {
            for(int i = 0; i < n; i++) scanf("%d", &val[i]), val[i]++;
            val[n] = 0;
            da(val, sa, n + 1, 1e6 + 5);
            calheight(val, sa, n);
            int l = 1, r = n, ans = 0;
            while(l <= r) {
                int mid = (l + r) >> 1;
                if(ok(mid)) {
                    ans = mid; l = mid + 1;
                }
                else r = mid - 1;
            }
            printf("%d
    ", ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/rolight/p/3993941.html
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