zoukankan      html  css  js  c++  java
  • LightOJ 1094

    Description:

    Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum amongst all nodes.

    Input:

    Input starts with an integer T (≤ 10), denoting the number of test cases.

    Each case starts with an integer n (2 ≤ n ≤ 30000) denoting the total number of nodes in the tree. The nodes are numbered from 0 to n-1. Each of the next n-1 lines will contain three integers u v w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 10000) denoting that node u and v are connected by an edge whose weight is w. You can assume that the input will form a valid tree.

    Output:

    For each case, print the case number and the maximum distance.

    Sample Input:

    2

    4

    0 1 20

    1 2 30

    2 3 50

    5

    0 2 20

    2 1 10

    0 3 29

    0 4 50

    Sample Output:

    Case 1: 100

    Case 2: 80

    题意:有n个点,它们之间连接的边有权值,问两个点之间最大的权值是多少。(DFS或BFS求树的直径)

    1.DFS:

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    
    const int N=100010;
    
    struct node
    {
        int v, weight, num;
    }no[N];
    int head[N], Max, id, dist[N], k; ///Max保存最大的权值,id保存最大权值的终点,dist数组保存以每个点为终点的边的权值
    
    void Add(int a, int b, int c) ///建立邻接表
    {
        no[k].v = b;
        no[k].num = head[a];
        no[k].weight = c;
    
        head[a] = k++;
    }
    
    void DFS(int u, int w)
    {
        int i, v;
    
        dist[u] = w;
    
        if (dist[u] > Max) ///更新最大权值及其终点
        {
            Max = dist[u];
            id = u;
        }
    
        for (i = head[u]; i != -1; i = no[i].num)
        {
            v = no[i].v;
    
            if (dist[v] == -1)
                DFS(v, dist[u]+no[i].weight);
        }
    }
    
    int main ()
    {
        int T, n, i, a, b, c, cnt = 0;
    
        scanf("%d", &T);
    
        while (T--)
        {
            cnt++;
    
            memset(head, -1, sizeof(head));
            k = 0;
    
            scanf("%d", &n);
            for (i = 1; i < n; i++)
            {
                scanf("%d%d%d", &a, &b, &c);
    
                Add(a, b, c);
                Add(b, a, c);
            }
    
            Max = 0;
            memset(dist, -1, sizeof(dist));
            DFS(0, 0); ///第一次查找后找到一个最大值及其终点
            memset(dist, -1, sizeof(dist));
            DFS(id, 0); ///第二次查找以上一次查找的终点开始
    
            printf("Case %d: %d
    ", cnt, Max);
        }
    
        return 0;
    }

    2.BFS:

    #include<stdio.h>
    #include<string.h>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    const int N=100010;
    
    struct node
    {
        int v, weight, num;
    }no[N];
    int dist[N], head[N], k, Max, id;
    
    void Add(int a, int b, int c)
    {
        no[k].v = b;
        no[k].weight = c;
        no[k].num = head[a];
    
        head[a] = k++;
    }
    
    void BFS(int ans)
    {
        int i, u, v;
        queue<int>Q;
    
        Q.push(ans);
        dist[ans] = 0;
    
        while (!Q.empty())
        {
            u = Q.front(); Q.pop();
    
            if (dist[u] > Max)
            {
                Max = dist[u];
                id = u;
            }
    
            for (i = head[u]; i != -1; i = no[i].num)
            {
                v = no[i].v;
    
                if (dist[v] == -1)
                {
                    dist[v] = dist[u] + no[i].weight;
                    Q.push(v);
                }
            }
        }
    }
    
    int main ()
    {
        int T, n, i, a, b, c, cnt = 0;
    
        scanf("%d", &T);
    
        while (T--)
        {
            cnt++;
            
            k = 0;
            memset(head, -1, sizeof(head));
    
            scanf("%d", &n);
            for (i = 1; i < n; i++)
            {
                scanf("%d%d%d", &a, &b, &c);
    
                Add(a, b, c);
                Add(b, a, c);
            }
    
            Max = 0;
            memset(dist, -1, sizeof(dist));
            BFS(0);
            memset(dist, -1, sizeof(dist));
            BFS(id);
    
            printf("Case %d: %d
    ", cnt, Max);
        }
    
        return 0;
    }
  • 相关阅读:
    CCNET的部分配置使用说明
    用Visual C++实现PDF文件的显示
    开发流程中的可用性
    TCP/IP学习笔记(一)
    使用 Microsoft 实时通信 API 增强多客户端通信
    TCP/IP学习笔记(二)IP网际协议
    DDE热链接
    软件设计中的可用性
    封包技术
    使用MFC开发ActiveX控件
  • 原文地址:https://www.cnblogs.com/syhandll/p/4739387.html
Copyright © 2011-2022 走看看