Codeforces 14D Two Paths 【树的直径】

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题目链接：Codeforces 14D Two Paths

题意：给你一棵树，让你找到两条不存在公共点的路径，记长度分别为x和y，求最大的x∗y。

思路：为了保证不存在公共点，我们划分两棵树，划分树的时候直接枚举边就可以了。对每一棵树求一次树的直径，维护最大值就好了。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#define PI acos(-1.0)#define CLR(a, b) memset(a, (b), sizeof(a))#define fi first#define se second#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 2*1e5+100;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;void getmax(int &a, int b) {a = max(a, b); }void getmin(int &a, int b) {a = min(a, b); }void add(LL &x, LL y) { x += y; x %= MOD; }struct Edge{    int from, to, next;};Edge edge[200*210];int head[210], edgenum;void init() {edgenum = 0; CLR(head, -1);}void addEdge(int u, int v) {    Edge E = {u, v, head[u]};    edge[edgenum] = E;    head[u] = edgenum++;}int node, ans;int dist[210];bool vis[210];void BFS(int sx){    queue<int> Q;    CLR(dist, 0); CLR(vis, false);    vis[sx] = true;    ans = 0; node = sx; Q.push(sx);    while(!Q.empty())    {        int u = Q.front(); Q.pop();        for(int i = head[u]; i != -1; i = edge[i].next)        {            Edge E = edge[i];            if(!vis[E.to] && dist[E.to] < dist[u] + 1)            {                vis[E.to] = true;                dist[E.to] = dist[u] + 1;                if(dist[E.to] > ans)                {                    ans = dist[E.to];                    node = E.to;                }                Q.push(E.to);            }        }    }}int Solve(int x){    BFS(x); BFS(node);    return ans;}int u[210], v[210];int main(){    int n; cin >> n;    for(int i = 1; i < n; i++) {        cin >> u[i] >> v[i];    }    int ans = 0;    for(int i = 1; i < n; i++) {        init();        for(int j = 1; j < n; j++) {            if(i == j) continue;            addEdge(u[j], v[j]);            addEdge(v[j], u[j]);        }        int x = Solve(u[i]);        int y = Solve(v[i]);        //cout << x << " " << y << endl;        ans = max(ans, x * y);    }    cout << ans << endl;    return 0;}

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