标签:nes 有一个 div with ++ more algo via a*
https://www.luogu.org/problem/show?pid=3565
枚举中间点
枚举中间点的子树
枚举距离
如果只有3个子树,那么对答案的贡献为a*b*c
假设现在来了第4个子树,那么答案会增加 d*(a*b+a*c+b*c)
再来第5个,答案增加e*(a*b+a*c+a*d+b*c+b*d+c*d)
括号里的数好像有点儿规律
用c2维护来维护它
d*(a*b+a*c+b*c)变到 e*(a*b+a*c+a*d+b*c+b*d+c*d)
增加了 e*(a+b+c)
所以用再用c1维护 前缀和
#include<cstdio> #include<cstring> #include<algorithm> #define N 5001 using namespace std; int front[N],nxt[N<<1],to[N<<1],tot; int dep[N],c1[N]; int maxd; long long ans,c2[N]; void add(int u,int v) { to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; to[++tot]=u; nxt[tot]=front[v]; front[v]=tot; } void dfs(int x,int f,int d) { maxd=max(maxd,d); dep[d]++; for(int i=front[x];i;i=nxt[i]) if(to[i]!=f) dfs(to[i],x,d+1); } int main() { int n,u,v; scanf("%d",&n); for(int i=1;i<n;i++) scanf("%d%d",&u,&v),add(u,v); for(int i=1;i<=n;i++) { for(int j=front[i];j;j=nxt[j]) { maxd=1; dfs(to[j],i,1); for(int k=1;k<=maxd;k++) { ans+=dep[k]*c2[k]; c2[k]+=c1[k]*dep[k]; c1[k]+=dep[k]; dep[k]=0; } } memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); } printf("%lld",ans); }
There are nn towns in Byteotia, connected with only n-1n−1 roads.
Each road directly links two towns.
All the roads have the same length and are two way.
It is known that every town can be reached from every other town via a route consisting of one or more (direct-link) roads.
In other words, the road network forms a tree.
Byteasar, the king of Byteotia, wants three luxury hotels erected to attract tourists from all over the world.
The king desires that the hotels be in different towns and at the same distance one from each other.
Help the king out by writing a program that determines the number of possible locations of the hotel triplet in Byteotia.
有一个树形结构,每条边的长度相同,任意两个节点可以相互到达。选3个点。两两距离相等。有多少种方案?
输入格式:
The first line of the standard input contains a single integer nn (1\le n\le 5\ 0001≤n≤5 000), the number of towns in Byteotia.
The towns are numbered from 11 to nn.
The Byteotian road network is then described in n-1n−1 lines.
Each line contains two integers aa and bb (1\le a\le b\le n1≤a≤b≤n) , separated by a single space, that indicate there is a direct road between the towns aa and bb.
输出格式:
The first and only line of the standard output should contain a single integer equal to the number of possible placements of the hotels.
7 1 2 5 7 2 5 2 3 5 6 4 5
5
有一个树形结构,每条边的长度相同,任意两个节点可以相互到达。选3个点。两两距离相等。有多少种方案?
标签:nes 有一个 div with ++ more algo via a*
原文地址:http://www.cnblogs.com/TheRoadToTheGold/p/7512508.html