Codeforces 14D Two Paths 树的直径

题目链接：点击打开链接

题意：给定一棵树

找2条点不反复的路径，使得两路径的长度乘积最大

思路：

1、为了保证点不反复，在图中删去一条边，枚举这条删边

2、这样得到了2个树，在各自的树中找最长链。即树的直径，然后相乘就可以

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<set>
#include<vector>
#include<map>
#include<math.h>
#include<queue>
#include<string>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define N 220
struct Edge {
int from, to, nex;
bool hehe;
}edge[N<<1];
int head
, edgenum;
void add(int u, int v){
Edge E={u,v,head[u],true};
edge[edgenum] = E;
head[u] = edgenum ++;
}
int n;
int dis
;
void init(){ memset(head, -1, sizeof head);	edgenum = 0; }
int bfs(int x){
memset(dis, 0, sizeof dis);
dis[x] = 1;
queue<int>q; q.push(x);
int hehe = x;
while(!q.empty()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = edge[i].nex) {
if(!edge[i].hehe)continue;
int v = edge[i].to;
if(dis[v])continue;
dis[v] = dis[u]+1, q.push(v);
if(dis[hehe]<dis[v])hehe = v;
}
}
return hehe;
}
int main(){
int i, u, v;
while(~scanf("%d",&n)){
init();
for(i=1;i<n;i++){
scanf("%d %d",&u,&v);
add(u,v); add(v,u);
}
int ans = 0;
for(i = 0; i < edgenum; i+=2)
{
edge[i].hehe = edge[i^1].hehe = false;
u = edge[i].from; v = edge[i].to;
int L1 = bfs(u); int R1 = bfs(L1);
int mul1 = dis[R1] -1;
int L2 = bfs(v); int R2 = bfs(L2);
int mul2 = dis[R2] -1;
ans = max(ans, mul1 * mul2);
edge[i].hehe = edge[i^1].hehe = true;
}
printf("%d\n",ans);
}
return 0;
}