• 【UOJ #35】后缀排序 后缀数组模板


    http://uoj.ac/problem/35
    以前做后缀数组的题直接粘模板。。。现在重新写一下模板
    注意用来基数排序的数组一定要开到N。

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int N = 100003;
    
    int t1[N], t2[N], c[N];
    
    void st(int *x, int *y, int *sa, int n, int m) {
    	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];
    }
    
    void mkhz(int *r, int *sa, int n, int m) {
    	int *x, *y, *t, i, j, p;
    	x = t1; y = t2;
    	for (i = 0; i < n; ++i) x[i] = r[i], y[i] = i;
    	st(x, y, sa, n, m);
    	for (p = 1, j = 1; j < n && p < n; j <<= 1, m = p - 1) {
    		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;
    		st(x, y, sa, n, m);
    		for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, 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++;
    	}
    }
    
    void mkh(int *r, int *sa, int *rank, int *h, int n) {
    	h[1] = 0; int k = 0, j;
    	for (int i = 1; i < n; ++i) rank[sa[i]] = i;
    	for (int i = 1; i < n; h[rank[i++]] = k)
    		for (k ? --k : k = 0, j = sa[rank[i] - 1]; r[j + k] == r[i + k]; ++k);
    }
    
    char s[N];
    int r[N], sa[N], h[N], rank[N];
    
    int main() {
    	scanf("%s", s + 1);
    	int len = strlen(s + 1);
    	for (int i = 1; i <= len; ++i) r[i] = s[i] - 'a' + 1;
    	mkhz(r, sa, len + 1, 26);
    	for (int i = 1; i <= len; ++i) printf("%d ", sa[i]);
    	puts("");
    	
    	mkh(r, sa, rank, h, len + 1);
    	for (int i = 2; i <= len; ++i) printf("%d ", h[i]);
    	puts("");
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/abclzr/p/6171089.html
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