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  • [TJOI2015]弦论(后缀自动机)

    /*
    一道在树上乱搞的题目
    建立出parent树来, 然后就能搞出每个节点往后能扩展出几个串, 至于位置不同算同一个的话就强制让right集合大小为1即可
    然后在树上类比权值线段树找第k大26分统计一下即可 
    
    */
    
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<queue>
    #include<iostream>
    #define ll long long 
    #define mmp make_pair
    #define M 1000100
    using namespace std;
    int read()
    {
    	int nm = 0, f = 1;
    	char c = getchar();
    	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    	return nm * f;
    }
    int t, k;
    char s[M];
    int ch[M][26], sz[M], len[M], fa[M], tim[M], a[M], f[M], lst = 1, cnt = 1;
    
    void insert(int c)
    {
    	int p = ++cnt, f = lst;
    	lst = p;
    	len[p] = len[f] + 1;
    	sz[p] = 1;
    	while(f && !ch[f][c]) ch[f][c] = p, f = fa[f];
    	if(!f) fa[p] = 1;
    	else
    	{
    		int q = ch[f][c];
    		if(len[q] == len[f] + 1) fa[p] = q;
    		else
    		{
    			int nq = ++cnt;
    			memcpy(ch[nq], ch[q], sizeof(ch[q]));
    			fa[nq] = fa[q];
    			len[nq] = len[f] + 1;
    			fa[q] = fa[p] = nq;
    			while(f && ch[f][c] == q) ch[f][c] = nq, f = fa[f];
    		}
    	}
    }
     
    void query(int now, int k)
    {
    	if(k <= sz[now]) return;
    	k -= sz[now];
    	for(int i = 0; i < 26; i++)
    	{
    		if(f[ch[now][i]] < k) k -= f[ch[now][i]];
    		else
    		{
    			putchar('a' + i);
    			query(ch[now][i], k);
    			break;
    		}
    	}
    }
     
    int main()
    {
    	scanf("%s", s + 1);
    	int l = strlen(s + 1);
    	for(int i = 1; i <= l; i++) insert(s[i] - 'a');
    	for(int i = 1; i <= cnt; i++) tim[len[i]]++;
    	for(int i = 1; i <= cnt; i++) tim[i] += tim[i - 1];
    	for(int i = 1; i <= cnt; i++) a[tim[len[i]]--] = i;
    	for(int i = cnt; i >= 1; i--) sz[fa[a[i]]] += sz[a[i]];
    	t = read(), k = read();
    	if(t == 0) for(int i = 1; i <= cnt; i++) f[i] = sz[i] = 1;
    	else for(int i = 1; i <= cnt; i++) f[i] = sz[i];
    	f[1] = sz[1] = 0;
    	for(int i = cnt; i >= 1; i--)
    	{
    		for(int j = 0; j < 26; j++)
    		{
    			f[a[i]] += f[ch[a[i]][j]];
    		}
    	}
    	if(k > f[1]) return 0 * puts("-1");
    	else query(1, k);	
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/luoyibujue/p/10628739.html
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