WOJ1024 (POJ1985+POJ2631) Exploration 树/BFS

---

title: WOJ1024 (POJ1985+POJ2631) Exploration 树/BFS

date: 2020-03-20 10:43:00

categories: acm

tags: [acm,woj,图论]

用到了树的 直径

1 描述

Tom is an explorer and now he is in a mysterious cave. He finds that there are lots of rooms in it. You are ensured that these rooms are connected together by some roads and there is only one way between any two rooms. That is, all the rooms are connected as a tree.

Now, Tom wants to travel all the rooms in this mysterious cave. The roads and rooms can be passed more than once. But there are too many rooms here and he doesn?t want to waste too much time and wants to find a walking sequence which can minimum the distance he walks. Note, Tom can select any room in the cave as the start point in his exploration.

Given the distance of roads in the cave, please write a program to calculate the minimum distance Tom has to go in his exploration.

输入格式

Standard input will contain multiple test cases. The first line of the input if a single integer T (1 <= T <= 50) which is the number of test cases.

Each test case starts with an integer N (2 <= N <= 50000), which is the number of the rooms. The following N - 1 lines contains three integers each, u (1 <= u <= N), v (1 <= v <= N) and d (1 <= d <= 2000). u and v is the serial number of the rooms and d is the distance between Room u and Room v .Note that, the serial numbers of the rooms start at 1.

输出格式

Results should be directed to standard output. Start each case with "Case #:" on a single line, where # is the case number starting from 1. Two consecutive cases should be separated by a single blank line. No blank line should be produced after the last test case.

For each test case, print a line containing the minimum distance Tom should go.

样例输入

2

2

1 2 3

3

1 2 3

1 3 4

样例输出

Case 1:

3

Case 2:

7

2 分析

给一棵树(节点间只有一条路(可能无边)),节点间边有正权。求从任一节点开始遍历这棵树所有点的最小权（节点可重复经过）

笔记加一条 数据结构-树的性质。树性质是向上只有一个节点，向下有没有限制个节点。(好像没什么用)，

画图知道无论从哪个点开始沿一条边出发，把经过这条边可以到达的点看成集合A，整个树为集合T，要想去T-S，必然要原路返回到起始点

(就是说把这个点砍了就是二分图)。所以这题看成两个过程，先是从起始点s遍历A,然后从最后到的点返回到起始点s，再遍历T-S

然后再画图可知从哪个点开始都是一样的(大雾，不一样!)

然后我感觉做过这种题，搜了一下想起来了,只需要找到树的直径然后直径上其他的树枝走两遍就行了，直径只走一遍

然后转化为无根树求直径的问题.

两次DFS/BFS都可以。任选一点做DFS，找到的最长点再做DFS，再次找到的最长点就是直径

//当然最短路Bellmanford 等算法也可以

类似的题POJ1985+POJ2631

3 code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
struct Node{
    int v,d;
};

int n,dis[50001];
vector<Node> tree[50001];
Node tmp1,tmp2;
int T,casee=0;
int u,v,d,i,ans,faresta,tpans;

int BFS(int x)
{
	for(i=0;i<=n;i++) //n.输入点的index到了n,一开始顺手写成n-1了
        dis[i]=0;
    int vis[n+5];
	memset(vis,0,sizeof(vis));
	queue<int> q;
	q.push(x);
	vis[x]=1;
	int point = 1;
	while(!q.empty())
	{
		int f=q.front();
		q.pop();
        if(dis[f]>tpans)
		{
		    tpans = dis[f];
			point = f;
        }
		for(i=0;i<tree[f].size();i++)
		{
			Node tmp = tree[f][i];
			if(vis[tmp.v]==0)
			{
				vis[tmp.v]=1;
				dis[tmp.v] = dis[f] + tmp.d;
				q.push(tmp.v);
			}
		}
	}
	return point;
}

void build(){
    for(i=1;i<=n;i++)
        tree[i].clear();
        ans=0;
        for(i=0;i<n-1;i++){
            cin>>u>>v>>d;
            tmp1.v=v;tmp1.d=d;
            tmp2.v=u;tmp2.d=d;
            tree[u].push_back(tmp1);  //相连的边,点
            tree[v].push_back(tmp2);
            ans=ans+2*d;
        }
    return;
}

int main(){

    cin>>T;
    while(T--){
        ans=0;
        if(casee!=0)
            cout<<endl;
        casee++;
        cin>>n;
        build();

        tpans=0;
        faresta=BFS(1);
        tpans=0;
        BFS(faresta);
        printf("Case %d:\n",casee);
        cout<<ans-tpans<<endl;
    }
    system("pause");
    return 0;
}