zoukankan      html  css  js  c++  java
  • HDU5313——DP+vector——Bipartite Graph

    Soda has a bipartite graph with n vertices and m undirected edges. Now he wants to make the graph become a complete bipartite graph with most edges by adding some extra edges. Soda needs you to tell him the maximum number of edges he can add.

    Note: There must be at most one edge between any pair of vertices both in the new graph and old graph.

     


    Input
    There are multiple test cases. The first line of input contains an integer T (1T100), indicating the number of test cases. For each test case:

    The first line contains two integers n and m(2n10000,0m100000).

    Each of the next m lines contains two integer u,v (1u,vn,vu) which means there's an undirected edge between vertex u and vertex v.

    There's at most one edge between any pair of vertices. Most test cases are small.
     


    Output
    For each test case, output the maximum number of edges Soda can add.
     


    Sample Input
    2 4 2 1 2 2 3 4 4 1 2 1 4 2 3 3 4
     


    Sample Output
    2 0
     


    Source
     


    Recommend
    /*
    大意:找到最长的上升序列(要求连在一起)
    DP思想 dp[u] += dp[v]
    从最长的开始找上升
    */
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    using namespace std;
    int n;
    int b[500010];
    vector<int> G[500010];
    int dp[500010];
    struct edge{
        int num, id;
    }a[500010];
    
    bool cmp(edge i, edge j)
    {
        return i.num < j.num;
    }
    int main()
    {
        int x, y;
       while(~scanf("%d", &n)){
           for(int i = 1; i < n ; i++)
               G[i].clear();
        for(int i = 1; i <= n ; i++){
            scanf("%d", &a[i].num);
            a[i].id = i;
        }
        for(int i = 1; i <= n; i++)
            b[i] = a[i].num;
        sort(a + 1, a + n + 1,cmp);
        for(int i = 1; i < n ; i++){
            scanf("%d%d", &x, &y);
            G[y].push_back(x);
            G[x].push_back(y);
        }
        int max1 = 1;
        memset(dp, 0, sizeof(dp));
        for(int i = n ; i >= 1; i--){
            int u = a[i].id;
            dp[u] = 1;
            for(int j = 0 ; j < G[u].size(); j++){
                int  v = G[u][j];
                if(b[v] > b[u]) {
                    dp[u] += dp[v];
                   // printf("%d
    ", dp[u]);
            }
            }
            max1 = max(dp[u], max1);
        }
        printf("%d
    ", max1);
       }
       return 0;
    }
    

      

  • 相关阅读:
    C++右值引用的参考
    U3D 文档 GPU INSTANCING
    UNITY statistic中的 SetPass和Batches
    时间复杂度
    转,数组遍历的三种方式
    bug纪录:PhotonServer-14052: 17:14:09.033
    关于.net standard 与 .net core, net framework
    【转】未能加载文件或程序集或它的某一个依赖项,系统找不到指定的文件
    C# 计时函数精度测试
    一张图看懂dex
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4694185.html
Copyright © 2011-2022 走看看