zoukankan      html  css  js  c++  java
  • POJ——T2271 Guardian of Decency

    http://poj.org/problem?id=2771

    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 5932   Accepted: 2463

    Description

    Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of them might become couples. While you can never exclude this possibility, he has made some rules that he thinks indicates a low probability two persons will become a couple: 

    • Their height differs by more than 40 cm. 
    • They are of the same sex. 
    • Their preferred music style is different. 
    • Their favourite sport is the same (they are likely to be fans of different teams and that would result in fighting).


    So, for any two persons that he brings on the excursion, they must satisfy at least one of the requirements above. Help him find the maximum number of persons he can take, given their vital information. 

    Input

    The first line of the input consists of an integer T ≤ 100 giving the number of test cases. The first line of each test case consists of an integer N ≤ 500 giving the number of pupils. Next there will be one line for each pupil consisting of four space-separated data items: 
    • an integer h giving the height in cm; 
    • a character 'F' for female or 'M' for male; 
    • a string describing the preferred music style; 
    • a string with the name of the favourite sport.

    No string in the input will contain more than 100 characters, nor will any string contain any whitespace. 

    Output

    For each test case in the input there should be one line with an integer giving the maximum number of eligible pupils.

    Sample Input

    2
    4
    35 M classicism programming
    0 M baroque skiing
    43 M baroque chess
    30 F baroque soccer
    8
    27 M romance programming
    194 F baroque programming
    67 M baroque ping-pong
    51 M classicism programming
    80 M classicism Paintball
    35 M baroque ping-pong
    39 F romance ping-pong
    110 M romance Paintball
    

    Sample Output

    3
    7
    

    Source

     
     1 #include <algorithm>
     2 #include <iostream>
     3 #include <cstring>
     4 #include <cstdio>
     5 
     6 using namespace std;
     7 
     8 const int N(500+15);
     9 int t,n,h,ans;
    10 
    11 int sumb,sumg;
    12 struct Node
    13 {
    14     int heigh;
    15     char music[119],sport[119];
    16 }boy[N],girl[N];
    17 
    18 int head[N],sumedge;
    19 struct Edge
    20 {
    21     int u,v,next;
    22     Edge(int u=0,int v=0,int next=0):
    23         u(u),v(v),next(next){}
    24 }edge[N*N];
    25 void ins(int u,int v)
    26 {
    27     edge[++sumedge]=Edge(u,v,head[u]);
    28     head[u]=sumedge;
    29 }
    30 
    31 int vis[N],match[N];
    32 bool DFS(int x)
    33 {
    34     for(int i=head[x];i;i=edge[i].next)
    35     {
    36         int v=edge[i].v;
    37         if(vis[v]) continue;
    38         vis[v]=1;
    39         if(!match[v]||DFS(match[v]))
    40         {
    41             match[v]=x;
    42             return true;
    43         }
    44     }
    45     return false;
    46 }
    47 
    48 void init()
    49 {
    50     ans=sumedge=sumb=sumg=0;
    51     memset(boy,0,sizeof(boy));
    52     memset(girl,0,sizeof(girl));
    53     memset(head,0,sizeof(head));
    54     memset(match,0,sizeof(match));
    55     scanf("%d",&n);
    56 }
    57 
    58 int main()
    59 {
    60     scanf("%d",&t);
    61     for(;t--;)
    62     {
    63         init();char xb;
    64         for(int h,i=1;i<=n;i++)
    65         {
    66             scanf("%d",&h);cin>>xb;
    67             if(xb=='F') boy[++sumb].heigh=h,scanf("%s%s",boy[sumb].music,boy[sumb].sport);
    68             else girl[++sumg].heigh=h,scanf("%s%s",girl[sumg].music,girl[sumg].sport);
    69         }
    70         for(int i=1;i<=sumb;i++)
    71           for(int j=1;j<=sumg;j++)
    72             if(abs(boy[i].heigh-girl[j].heigh)<=40&&!strcmp(boy[i].music,girl[j].music)&&strcmp(boy[i].sport,girl[j].sport))
    73              ins(i,j);
    74         for(int i=1;i<=sumb;i++)
    75         {
    76             memset(vis,0,sizeof(vis));
    77             if(DFS(i)) ans++;
    78         }
    79         printf("%d
    ",n-ans);
    80     }
    81     return 0;
    82 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
  • 相关阅读:
    Blue的博客
    透明状态栏和沉浸式状态栏
    Html的label和span的区别
    三个石匠的故事
    OpenSSL 生成自定义证书
    github博客配置
    js作用域其二:预解析
    数据分析常用工具总结
    堆排序
    吴裕雄--天生自然 JAVA开发学习: 循环结构
  • 原文地址:https://www.cnblogs.com/Shy-key/p/7113177.html
Copyright © 2011-2022 走看看