Cotree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1215 Accepted Submission(s): 416
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 ∑ni=1∑nj=i+1dis(i,j), where dis(i,j) represents the number of edges of the path from i to j. He is happy with only the function value.
Input
The first line contains a number n (2<=n<=100000). In each of the following n−2 lines, there are two numbers u and v, meaning that there is an edge between u and v. The input is guaranteed to contain exactly two trees.
Output
Just print the minimum function value.
Sample Input
3
1 2
Sample Output
4
Source
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liuyiding
题意：给定两棵树，然后让你加上一条边使得成为一棵树，并且新树上的所有的任意两点的距离最小。
析：首先根据题意应该能够知道连接两棵树的重心才是最小的距离，树的重心查找方式就是枚举每个点，然后计算去年该点剩下的连通分量中点的数量最多的需要最少，只需要一个DFS就可以解决。有了重心后，可以直接连一条边，然后再计算距离，距离计算可以枚举所有边的贡献，该边的贡献就是该边左点的点数乘以该边右边的点数。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define print(x) cout<<(x)<<endl
// #define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
// #define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define FOR(i,n,x) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 7;
const int maxm = 2000000 + 7;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
int n, m;
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){
int x; cin >> x; return x;
}
struct Edge{
int to, next;
};
int head[maxn];
int cnt;
Edge edges[maxn<<1];
int f[maxn], sz[maxn];
int p[maxn];
inline int Find(int x){
return x == p[x] ? x : p[x] = Find(p[x]);
}
void add_edge(int u, int v){
edges[cnt].to = v;
edges[cnt].next = head[u];
head[u] = cnt++;
}
void dfs(int u, int fa, int num, int &rt){
sz[u] = 1;
f[u] = 0;
for(int i = head[u]; ~i; i = edges[i].next){
int v = edges[i].to;
if(v != fa){
dfs(v, u, num, rt);
sz[u] += sz[v];
f[u] = max(f[u], sz[v]);
}
}
f[u] = max(f[u], num - sz[u]);
if(f[u] < f[rt]) rt = u;
}
int dp[maxn];
LL dfs(int u, int fa){
dp[u] = 1;
LL ans = 0;
for(int i = head[u]; ~i; i = edges[i].next){
int v = edges[i].to;
if(v != fa){
ans += dfs(v, u);
dp[u] += dp[v];
ans += dp[v] * 1LL * (n - dp[v]);
}
}
return ans;
}
map<int, int> mp;
int main(){
while(scanf("%d", &n) == 1){
cnt = 0; ms(head, -1);
for(int i = 0; i <= n; ++i) p[i] = i;
for(int i = 2; i < n; ++i){
int x, y; scanf("%d %d", &x, &y);
add_edge(x, y);
add_edge(y, x);
x = Find(x);
y = Find(y);
if(x != y) p[y] = x;
}
mp.cl;
for(int i = 1; i <= n; ++i) ++mp[Find(i)];
ms(f, INF);
auto it = mp.be;
int rt1 = 0, rt2 = 0;
dfs(it->fi, -1, it->se, rt1);
++it;
dfs(it->fi, -1, it->se, rt2);
add_edge(rt1, rt2);
add_edge(rt2, rt1);
print(dfs(1, -1));
}
return 0;
}