UVA 11695 Flight Planning 修改一条边使得树的直径最短

题目链接：点击打开链接

题意：

给定n(n<=2500) 节点的一棵树

删除一条边再加入一条边使得树的直径最短。

思路：首先枚举删除的那条边，

然后计算出删除后的2棵子树各自的重心

则新建的边一定是重心的连线。

而新的直径要么是在某个子树中，要么是两个子树间。

#include <cstdio>#include <algorithm>#include <string.h>#include <queue>#include <cstring>#include <cmath>#include <iostream>#include <vector>using namespace std;#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;#define pb push_backconst double inf = 1e9;const int N = 2505;struct Ans{int ans, oldx, oldy, nowx, nowy;};vector<int>edge[N];void add(int u, int v){edge[u].push_back(v);}int n;void init(){ for (int i = 0; i <= n; i++)edge[i].clear(); }Ans a;int dis[N], pre[N], far, badu, badv;vector<int>G;void bfs(int x){for (int i = 1; i <= n; i++)dis[i] = -1;pre[x] = -1;dis[x] = 0;queue<int>q;q.push(x);far = x;while (!q.empty()){int u = q.front(); q.pop();for (int i = 0; i < edge[u].size(); i++){int v = edge[u][i];if (dis[v] != -1 || (min(u, v) == min(badu, badv) && max(u, v) == max(badu, badv)))continue;dis[v] = dis[u] + 1;pre[v] = u;q.push(v); if (dis[far] < dis[v])far = v;}}G.clear();int tmp = far;while (tmp != -1){G.pb(tmp); tmp = pre[tmp];}}void work(){Ans d = {0, badu, badv, 0, 0 };int link = 1;bfs(badu);bfs(far);link += (dis[far] + 1) / 2;d.nowx = G[G.size() / 2];d.ans = dis[far];bfs(badv);bfs(far);d.nowy = G[G.size() / 2];link += (dis[far] + 1) / 2;d.ans = max(d.ans, dis[far]);d.ans = max(d.ans, link);if (d.ans < a.ans)a = d;}int l[N], r[N];void input(){scanf("%d", &n);init();for (int i = 1, u, v; i < n; i++){scanf("%d %d", &u, &v);add(u, v); add(v, u);l[i] = u; r[i] = v;}a.ans = inf;}int main(){int T; scanf("%d", &T);while (T--){input();for (int i = 1; i < n; i++){badu = l[i]; badv = r[i];work();}printf("%d\n%d %d\n%d %d\n", a.ans, a.oldx, a.oldy, a.nowx, a.nowy);}return 0;}/*241 22 33 4141 21 82 32 48 98 108 114 54 64 710 1210 1313 14*/