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  • Bzoj2780: [Spoj]8093 Sevenk Love Oimaster

    题目

    传送门

    Sol

    就是广义(sam)
    然后记录下每个状态属于哪些串,开(set)维护
    (parent)树上启发式合并一下就好了

    # 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(2e5 + 5);
    
    int trans[26][maxn], fa[maxn], len[maxn], size[maxn], last, tot = 1;
    int l[maxn], r[maxn], n, q, t[maxn], id[maxn];
    set <int> :: iterator it;
    set <int> ed[maxn];
    char s[maxn], qs[maxn << 1];
    
    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[q] = fa[np] = nq;
    			while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
    		}
    	}
    }
    
    int main(){
    	n = Input(), q = Input();
    	for(RG int i = 1; i <= n; ++i){
    		l[i] = r[i - 1];
    		scanf(" %s", s + l[i]);
    		r[i] = l[i] + strlen(s + l[i]), last = 1;
    		for(RG int j = l[i]; j < r[i]; ++j) Extend(s[j] - 'a');
    	}
    	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 = 1; i <= n; ++i)
    		for(RG int j = l[i], nw = 1; j < r[i]; ++j){
    			nw = trans[s[j] - 'a'][nw];
    			ed[nw].insert(i);
    		}
    	for(RG int i = tot; i; --i){
    		RG int p = id[i];
    		size[p] = ed[p].size();
    		if(size[p] > ed[fa[p]].size()) swap(ed[p], ed[fa[p]]);
    		for(it = ed[p].begin(); it != ed[p].end(); ++it)
    			ed[fa[p]].insert(*it);
    	}
    	for(RG int i = 1; i <= q; ++i){
    		scanf(" %s", qs);
    		RG int nw = 1, l = strlen(qs);
    		for(RG int j = 0; j < l; ++j) nw = trans[qs[j] - 'a'][nw];
    		printf("%d
    ", size[nw]);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8940867.html
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