Codeforces Beta Round #14 D - Two Paths
D - Two PathsCrawling in process...Crawling failedTime Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
SubmitStatusPracticeCodeForces 14D
Description
As you know, Bob's brother lives in Flatland. In Flatland there are n cities, connected by n - 1 two-way roads. The cities are numbered from 1 to n. You can get from one city to another moving along the roads.
The «Two Paths» company, where Bob's brother works, has won a tender to repair two paths in Flatland. A path is a sequence of different cities, connected sequentially by roads. The company is allowed to choose by itself the paths to repair. The only condition they have to meet is that the two paths shouldn't cross (i.e. shouldn't have common cities).
It is known that the profit, the «Two Paths» company will get, equals the product of the lengths of the two paths. Let's consider the length of each road equals 1, and the length of a path equals the amount of roads in it. Find the maximum possible profit for the company.
Input
The first line contains an integer n (2 ≤ n ≤ 200), where n is the amount of cities in the country. The following n - 1 lines contain the information about the roads. Each line contains a pair of numbers of the cities, connected by the road ai, bi (1 ≤ ai, bi ≤ n).
Output
Output the maximum possible profit.
Sample Input
Input
4
1 2
2 3
3 4
Output
1
Input
7
1 2
1 3
1 4
1 5
1 6
1 7
Output
0
Input
6
1 2
2 3
2 4
5 4
6 4
Output
4
[cpp]
//枚举每一条边，其实相当于把这条边给砍掉。
//例如第三个样例，假如我们枚举到2,4这条边，我们把这条边给砍掉
//然后从2开始走求两个最大值，相加就是我们得到的一个path
//同样从4开始走求两个最大值，相加是我们得到的另一个path
//所以这两个值相乘就得到满足题意的profit
//枚举每一条边，我们可以得到最大的profit值
#include<iostream>
#include<cstdlib>
#include<vector>
#include<stdio.h>
using namespace std;
const int N=205;
vector<int>v[N];
int n;
int s;
int dfs(int x,int pre)
{
int maxn1=0;
int maxn2=0;
int sum=0;
for(int i=0;i<v[x].size();i++)
{
if(v[x][i]!=pre)
{
sum=max(dfs(v[x][i],x),sum);
if(s>maxn1)
{
maxn2=maxn1;
maxn1=s;
}
else
maxn2=max(maxn2,s);
}
}
sum=max(sum,maxn1+maxn2);
s=maxn1+1;
return sum;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
v[i].clear();
int maxn=0;
for(int i=0;i<n-1;i++)
{
int a,b;
scanf("%d%d",&a,&b);
v[a].push_back(b);
v[b].push_back(a);
}
for(int i=1;i<=n;i++)
for(int j=0;j<v[i].size();j++)//枚举边
{
int a=dfs(v[i][j],i);
int b=dfs(i,v[i][j]);
if(a*b>maxn)
maxn=a*b;
}
cout<<maxn<<endl;
}
}
补充：软件开发 , C++ ,
