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  • Bzoj3277:串

    题面

    传送门

    Sol

    广义(sam)
    每个(sam)的状态开(set)记录属于哪些串
    (parent)树上启发式合并(set)

    然后每个串就在上面走,通过不停地跳(parent)树的父亲节点保证大于等于(k),贡献就是(len)

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
     
    IL int Input(){
        RG int x = 0, z = 1; RG char c = getchar();
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
     
    const int maxn(1e5 + 5);
    const int maxm(5e5 + 5);
    
    int trans[26][maxm], fa[maxm], tot = 1, last, len[maxm];
    int id[maxm], t[maxm], n, k, size[maxm];
    set <int> ed[maxm];
    set <int> :: iterator it;
    string s[maxn];
    
    IL void Extend(RG int c){
    	RG int p = last, np = ++tot;
    	len[last = np] = len[p] + 1;
    	while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    	if(!p) fa[np] = 1;
    	else{
    		RG int q = trans[c][p];
    		if(len[q] == len[p] + 1) fa[np] = q;
    		else{
    			RG int nq = ++tot;
    			fa[nq] = fa[q], len[nq] = len[p] + 1;
    			for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
    			fa[np] = fa[q] = nq;
    			while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
    		}
    	}
    }
    
    int main(RG int argc, RG char *argv[]){
    	n = Input(), k = Input();
    	for(RG int i = 1; i <= n; ++i){
    		cin >> s[i]; last = 1;
    		for(RG int j = 0, l = s[i].size(); j < l; ++j) Extend(s[i][j] - 'a');
    	}
    	for(RG int i = 1; i <= n; ++i)
    		for(RG int j = 0, nw = 1, l = s[i].size(); j < l; ++j){
    			nw = trans[s[i][j] - 'a'][nw];
    			ed[nw].insert(i);
    		}
    	for(RG int i = 1; i <= tot; ++i) ++t[len[i]];
    	for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
    	for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
    	for(RG int i = tot; i; --i){
    		RG int x = id[i];
    		size[x] = ed[x].size();
    		if(size[x] > size[fa[x]]) swap(ed[x], ed[fa[x]]);
    		for(it = ed[x].begin(); it != ed[x].end(); ++it)
    			ed[fa[x]].insert(*it);
    	}
    	for(RG int i = 1; i <= n; ++i){
    		RG ll ans = 0, nw = 1;
    		for(RG int j = 0, l = s[i].size(); j < l; ++j){
    			nw = trans[s[i][j] - 'a'][nw];
    			while(nw != 1 && size[nw] < k) nw = fa[nw];
    			if(size[nw] >= k) ans += len[nw];
    		}
    		printf("%lld ", ans);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8933664.html
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