D - Two Paths Crawling 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] //枚舉每一條邊，其實相當於把這條邊給砍掉。 //例如第三個樣例，假如我們枚舉到2,4這條邊，我們把這條邊給砍掉 //然後從2開始走求兩個最大值，相加就是我們得到的一個path //同樣從4開始走求兩個最大值，相加是我們得到的另一個path //所以這兩個值相乘就得到滿足題意的profit //枚舉每一條邊，我們可以得到最大的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++)//枚舉邊 { int a=dfs(v[i][j],i); int b=dfs(i,v[i][j]); if(a*b>maxn) maxn=a*b; } cout<<maxn<<endl; } }
