zoukankan      html  css  js  c++  java
  • POJ3020 Antenna Placement —— 最大匹配 or 最小边覆盖

    题目链接:https://vjudge.net/problem/POJ-3020

    Antenna Placement
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9995   Accepted: 4939

    Description

    The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking reason why they got the job, is their discovery of a new, highly noise resistant, antenna. It is called 4DAir, and comes in four types. Each type can only transmit and receive signals in a direction aligned with a (slightly skewed) latitudinal and longitudinal grid, because of the interacting electromagnetic field of the earth. The four types correspond to antennas operating in the directions north, west, south, and east, respectively. Below is an example picture of places of interest, depicted by twelve small rings, and nine 4DAir antennas depicted by ellipses covering them. 
     
    Obviously, it is desirable to use as few antennas as possible, but still provide coverage for each place of interest. We model the problem as follows: Let A be a rectangular matrix describing the surface of Sweden, where an entry of A either is a point of interest, which must be covered by at least one antenna, or empty space. Antennas can only be positioned at an entry in A. When an antenna is placed at row r and column c, this entry is considered covered, but also one of the neighbouring entries (c+1,r),(c,r+1),(c-1,r), or (c,r-1), is covered depending on the type chosen for this particular antenna. What is the least number of antennas for which there exists a placement in A such that all points of interest are covered? 

    Input

    On the first row of input is a single positive integer n, specifying the number of scenarios that follow. Each scenario begins with a row containing two positive integers h and w, with 1 <= h <= 40 and 0 < w <= 10. Thereafter is a matrix presented, describing the points of interest in Sweden in the form of h lines, each containing w characters from the set ['*','o']. A '*'-character symbolises a point of interest, whereas a 'o'-character represents open space. 

    Output

    For each scenario, output the minimum number of antennas necessary to cover all '*'-entries in the scenario's matrix, on a row of its own.

    Sample Input

    2
    7 9
    ooo**oooo
    **oo*ooo*
    o*oo**o**
    ooooooooo
    *******oo
    o*o*oo*oo
    *******oo
    10 1
    *
    *
    *
    o
    *
    *
    *
    *
    *
    *
    

    Sample Output

    17
    5

    Source

    题解:

    1.首先为每个“*”编号。然后对于当前的“*”, 如果它的上面有“*”,则在这两个“*”之间连一条边,同理其他三个方向。

    2.利用匈牙利算法求出最大匹配数cnt,即表明最多有cnt个“*”可以与其他“*”共用,所以最少需要N-cnt个。

    3.其实此题求的就是最小边覆盖:最小边覆盖 = 结点数 - 最大匹配数

    代码如下:

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <string>
     6 #include <vector>
     7 #include <map>
     8 #include <set>
     9 #include <queue>
    10 #include <sstream>
    11 #include <algorithm>
    12 using namespace std;
    13 const int INF = 2e9;
    14 const int MOD = 1e9+7;
    15 const int MAXN = 1600+10;
    16 
    17 int N;
    18 char a[MAXN][MAXN];
    19 int M[MAXN][MAXN], id[MAXN][MAXN], link[MAXN];
    20 bool vis[MAXN];
    21 
    22 bool dfs(int u)
    23 {
    24     for(int i = 1; i<=N; i++)
    25     if(M[u][i] && !vis[i])
    26     {
    27         vis[i] = true;
    28         if(link[i]==-1 || dfs(link[i]))
    29         {
    30             link[i] = u;
    31             return true;
    32         }
    33     }
    34     return false;
    35 }
    36 
    37 int hungary()
    38 {
    39     int ret = 0;
    40     memset(link, -1, sizeof(link));
    41     for(int i = 1; i<=N; i++)
    42     {
    43         memset(vis, 0, sizeof(vis));
    44         if(dfs(i)) ret++;
    45     }
    46     return ret;
    47 }
    48 
    49 int main()
    50 {
    51     int T, n, m;
    52     scanf("%d", &T);
    53     while(T--)
    54     {
    55         scanf("%d%d", &n, &m);
    56         N = 0;
    57         memset(id, -1, sizeof(id));
    58         for(int i = 1; i<=n; i++)
    59         {
    60             scanf("%s", a[i]+1);
    61             for(int j = 1; j<=m; j++)
    62                 if(a[i][j]=='*')
    63                     id[i][j] = ++N;
    64         }
    65 
    66         memset(M, false, sizeof(M));
    67         for(int i = 1; i<=n; i++)
    68         for(int j = 1; j<=m; j++)
    69         {
    70             if(id[i][j]==-1) continue;
    71             if(j!=1 && id[i][j-1]!=-1) M[id[i][j]][id[i][j-1]] = true;
    72             if(j!=m && id[i][j+1]!=-1) M[id[i][j]][id[i][j+1]] = true;
    73             if(i!=1 && id[i-1][j]!=-1) M[id[i][j]][id[i-1][j]] = true;
    74             if(i!=n && id[i+1][j]!=-1) M[id[i][j]][id[i+1][j]] = true;
    75         }
    76 
    77         int cnt = hungary()/2;
    78         printf("%d
    ", N-cnt);
    79     }
    80 }
    View Code
  • 相关阅读:
    Brain network involved in autonomic functions 与自主功能相关的大脑网络
    Brief summary of classical components of ERP 事件相关成分(ERP)经典成分小结
    ICA & Percentage Variance Account For (PVAF)
    数据处理中白化Whitening的作用图解分析
    Loadings vs eigenvectors in PCA 主成分分析(PCA)中的负荷和特征向量
    主成分分析(PCA)和独立成分分析(ICA)相关资料
    Sketch of heart and QRS complex 心脏及QRS波群简图
    Brain Network visulation in EEG 脑电网络可视化
    Phase Locking Value (PLV) 神经信号的锁相值
    ubuntu16.04下的一些基本操作笔记
  • 原文地址:https://www.cnblogs.com/DOLFAMINGO/p/7818351.html
Copyright © 2011-2022 走看看