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  • JZOJ 4211. 【五校联考1day2】送你一颗圣诞树

    题目

    解析

    (Code)

    #include<cstdio>
    #include<map>
    #include<iostream>
    #include<cstring>
    using namespace std;
    typedef unsigned long long LL;
    const int N = 70;
    const LL P = 1e9 + 7;
    
    map<pair<LL , pair<LL , LL> > , LL> vis_to;
    map<pair<LL , LL> , LL> vis_all;
    
    struct node{
    	LL a , b , c , d , l , sz , f;
    }q[N];
    
    inline LL get(LL x , LL u , LL v)
    {
    	if (q[x].sz == 1 || u == v) return 0;
    	if (vis_to[make_pair(x , make_pair(u , v))]) return vis_to[make_pair(x , make_pair(u , v))];
    	LL res = 0 , a = q[x].a , b = q[x].b , c = q[x].c , d = q[x].d , l = q[x].l , sz = q[a].sz;
    	if (u > v) swap(u , v);
    	if (v <= sz) res = get(a , u , v);
    	else if (u > sz) res = get(b , u - sz , v - sz);
    	else res = (get(a , u , c) + l + get(b , v - sz , d)) % P;
    	vis_to[make_pair(x , make_pair(u , v))] = res;
    	return res;
    }
    
    inline LL all(LL x , LL u)
    {
    	if (q[x].sz == 1) return 0;
    	if (vis_all[make_pair(x , u)]) return vis_all[make_pair(x , u)];
    	LL res = 0 , a = q[x].a , b = q[x].b , c = q[x].c , d = q[x].d , l = q[x].l , sz = q[a].sz;
    	if (u <= sz) res = ((all(a , u) + all(b , d)) % P + q[b].sz % P * ((l + get(a , u , c)) % P) % P) % P;
    	else res = ((all(b , u - sz) + all(a , c)) % P + q[a].sz % P * ((l + get(b , u - sz , d)) % P) % P) % P;
    	vis_all[make_pair(x , u)] = res;
    	return res;
    }
    
    int main()
    {
    	int T;
    	scanf("%d" , &T);
    	for(; T; T--)
    	{
    		vis_to.clear() , vis_all.clear();
    		int n;
    		scanf("%d" , &n);
    		memset(q , 0 , sizeof q);
    		q[0] = node{0 , 0 , 0 , 0 , 0 , 1};
    		LL a , b , c , d , l;
    		for(register int i = 1; i <= n; i++)
    		{
    			scanf("%lld%lld%lld%lld%lld" , &a , &b , &c , &d , &l);
    			++c , ++d;
    			q[i] = node{a , b , c , d , l , q[a].sz + q[b].sz};
    			q[i].f = ((q[a].f + q[b].f) % P + q[a].sz % P * all(b , d) % P 
    				+ q[b].sz % P * all(a , c) % P + q[a].sz % P * (q[b].sz % P) % P * l % P) % P;
    			printf("%lld
    " , q[i].f);
    		}
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/leiyuanze/p/13498517.html
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