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.

Examples
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
我承認一開始被這個分數和她的標簽嚇著了。。
后來發現n最大才二百的時候，直接枚舉每條邊，刪掉這條邊之后形成兩棵樹。求這兩棵樹的直徑，更新答案。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=100+100; vector<int> p[maxx]; int dp[maxx][2]; int ans; int n;inline int dfs(int u,int f) {int sum=0;int max1=0,max2=0;for(int i=0;i<p[u].size();i++){if(p[u][i]!=f){sum=max(sum,dfs(p[u][i],u));if(ans>max1){max2=max1;max1=ans;}else max2=max(max2,ans);}}sum=max(sum,max1+max2);ans=max1+1;return sum; } int main() {int x,y;scanf("%d",&n);for(int i=1;i<n;i++){scanf("%d%d",&x,&y);p[x].push_back(y);p[y].push_back(x);}int ans=0;int _max=0;for(int i=1;i<=n;i++){for(int j=0;j<p[i].size();j++){ans=0;int _max1=dfs(i,p[i][j]);int _max2=dfs(p[i][j],i);_max=max(_max,_max1*_max2);}}cout<<_max<<endl; }

努力加油a啊，(^o)/~

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