2019江西省省赛 Cotree[换根dp]-CFANZ编程社区

【题目】

hdu6567

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=6567

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 336 Accepted Submission(s): 129

Problem Description

Avin has two trees which are not connected. He asks you to add an edge between them to make them connected while minimizing the function , where represents the number of edges of the path from to . He is happy with only the function value.

Input

The first line contains a number . In each of the following lines, there are two numbers and , meaning that there is an edge between and . The input is guaranteed to contain exactly two trees.

Output

Just print the minimum function value.

Sample Input

3 1 2

Sample Output

Source

2019CCPC-江西省赛（重现赛）- 感谢南昌大学

【题意】

给两个树，连一条边连接两颗树，问如何连接使得树上两两之间的距离和最小。

【思路】

刚开始想的是找两颗树的重心连接，一直wa就换成下面这个思路：设dp[u] 为从u节点出发到达u子树所有节点的距离和。

那么连接就是两颗树的最小u节点连接即可。

那么我们可以一次dfs预处理所有节点到达儿子、孙子节点的距离和。

然后通过换根dp就可以得到所有的dp[i]了。

连接ccurid1,curid2跑一遍dfs3计算总距离即可

【代码】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
vector<int>G[N];
int n;
ll sz[N],dp[N],ty[N];
int curid1,curid2,curid;

void dfs1(int u,int fa)
{
    sz[u]=1;
    ty[u]=1;
    for(int v:G[u])
    {
        if(v==fa) continue;
        dfs1(v,u);
        sz[u]+=sz[v];
        dp[u]=dp[u]+dp[v]+sz[v];
    }
}
ll ans;
void dfs2(int u,int fa)
{
    if(dp[u]<ans){
        ans=dp[u];
        curid=u;
    }
    ans=min(ans,dp[u]);
    for(int v:G[u])
    {
        if(v==fa) continue;
        ll tmp=dp[u]-dp[v]-sz[v];
        dp[v]+=tmp+sz[u]-sz[v];
        sz[v]+=sz[u]-sz[v];
        dfs2(v,u);
    }
}
void dfs3(int u,int fa)
{
    sz[u]=1;
    for(int v:G[u])
    {
        if(v==fa) continue;
        dfs3(v,u);
        sz[u]+=sz[v];
        ans+=sz[v]*(n-sz[v]);
    }
}
int main()
{
    cin>>n;
    for(int i=1;i<=n-2;++i)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    int ed;
    dfs1(1,-1);
    ans=0x3f3f3f3f3f3f3f3f;
    curid=1;
    dfs2(1,-1);//开始换根
    curid1=curid;
    for(int i=1;i<=n;++i) if(ty[i]==0){ed=i;break;}
    dfs1(ed,-1);
    ans=0x3f3f3f3f3f3f3f3f;
    curid=ed;
    dfs2(ed,-1);//开始换根
    curid2=curid;
    G[curid1].push_back(curid2);
    ans=0;
    memset(sz,0,sizeof(sz));
    dfs3(curid1,-1);
    printf("%lld\n",ans);
}