Two Paths CodeForces - 14D（暴力+树的直径）

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
我承认一开始被这个分数和她的标签吓着了。。
后来发现n最大才二百的时候，直接枚举每条边，删掉这条边之后形成两棵树。求这两棵树的直径，更新答案。
代码如下：

#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)/~

Two Paths CodeForces - 14D（暴力+树的直径）相关推荐

Codeforces 337D Book of Evil：树的直径【结论】
题目链接:http://codeforces.com/problemset/problem/337/D 题意: 给你一棵树,n个节点. 如果一个节点处放着"罪恶之书",那么它会影响 ...
Book of Evil（树的直径+思维）
Book of Evil Paladin Manao caught the trail of the ancient Book of Evil in a swampy area. This area ...
0x63.图论 - 树的直径与最近公共祖先
目录一.树的直径(Diameter) 1.树形DP求树的直径 2.两次BFS/DFS求树的直径 1.POJ 1985.Cow Marathon(DFS求树的直径模板题) 2.AcWing 350. ...
hdu 4607 Park Visit 求树的直径
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4607 Claire and her little friend, ykwd, are travelli ...
51 nod 1427 文明 (并查集 + 树的直径)
1427 文明题目来源: CodeForces 基准时间限制:1.5 秒空间限制:131072 KB 分值: 160 难度:6级算法题安德鲁在玩一个叫"文明"的游戏.大妈正在 ...
历届试题大臣的旅费（深搜树的直径）
问题描述很久以前,T王国空前繁荣.为了更好地管理国家,王国修建了大量的快速路,用于连接首都和王国内的各大城市. 为节省经费,T国的大臣们经过思考,制定了一套优秀的修建方案,使得任何一个大城市都能从首 ...
nssl1438-战略威慑【枚举,树的直径】
正题题目大意 nnn个点的无根树,求两条不相交的路径使它们长度之积最大. 解题思路我们暴力枚举第一条,然后求树的直径即可. codecodecode #include<cstdio> ...
【洛谷 P3304】[SDOI2013]直径（树的直径）
题目链接题意,求一棵树被所有直径经过的边的条数. 这题是我们8.25KS图论的最后一题,当时我果断打了暴力求所有直径然后树上差分统计的方法,好像有点小问题,boom0了. 考完改这题,改了好久,各种 ...
树的直径/重心学习笔记
树的直径 POJ2631 Roads in the North 题意:裸的直径题解:套模板,但是开始的时候,我的代码在第一次dfs循环的时候,没有考虑到,路长全为0的情况,而用来记录最远点的maxp ...

Two Paths CodeForces - 14D（暴力+树的直径）

Two Paths CodeForces - 14D（暴力+树的直径）相关推荐

最新文章

热门文章