zoukankan      html  css  js  c++  java
  • HDU 2444 The Accomodation of Students (判断是否是二分图,然后求最大匹配)

    The Accomodation of Students

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

    Description
    There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

    Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

    Calculate the maximum number of pairs that can be arranged into these double rooms.
     
    Input
    For each data set:
    The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

    Proceed to the end of file.

     
    Output
    If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
     
    Sample Input
    4 4
    1 2
    1 3
    1 4
    2 3
    6 5
    1 2
    1 3
    1 4
    2 5
    3 6
     
    Sample Output
    No
    3
    题意:有n个学生,m对熟人,问你能否把他们分为2部分,其中在同一部分的人不认识,如果能,再将他们分到寝室,只有相互认识的人才能分到同一个寝室,问最多需要多少个寝室
    分析:先判断是否为二分图,不是,就输出No,否则就求最大匹配。保存图用了两种方式。
    /* ****************************************************
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    using namespace std;
    
    const int MAXN = 210;
    struct Edge{
        int to,next;
    }edge[MAXN*MAXN];
    int head[MAXN],tol;
    
    int uN,vN;
    int linker[MAXN];
    bool used[MAXN];
    
    int col[MAXN];
    
    void init()
    {
        tol=0;
        memset(head,-1,sizeof(head));
    }
    
    void addedge(int u,int v)
    {
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
    }
    
    bool dfs(int u)
    {
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;
            if(!used[v]){
                used[v]=true;
                if(linker[v]==-1||dfs(linker[v])){
                    linker[v]=u;
                    return true;
                }
            }
        }
        return false;
    }
    
    int Hungary()
    {
        int res=0;
        int u;
        memset(linker,-1,sizeof(linker));
        for(u=1;u<=uN;u++){
            memset(used,false,sizeof(used));
            if(dfs(u)) res++;
        }
        return res;
    }
    
    bool Color(int u)
    {
        if(col[u]==-1) col[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;
            if(col[v]==-1){
                col[v]=col[u]^1;
                if(!Color(v)) return false;
            }
            if(col[v]==col[u]) return false;
        }
        return true;
    }
    
    int main()
    {
        int n,m;
        int u,v;
        while(scanf("%d%d",&n,&m)!=EOF){
            init();
            uN=vN=n;
            for(int i=1;i<=m;i++){
                scanf("%d%d",&u,&v);
                addedge(u,v);
                addedge(v,u);
            }
            bool flag=true;
            memset(col,-1,sizeof(col));
            for(int i=1;i<=n;i++)
                if(col[i]==-1) flag&=Color(i);
            if(!flag){
                printf("No
    ");
                continue;
            }
            else printf("%d
    ",Hungary()/2);
        }
        return 0;
    }
    ************************************************ */
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<vector>
    using namespace std;
    
    const int MAXN = 210;
    vector<int> G[MAXN];
    int uN;
    int linker[MAXN];
    bool used[MAXN];
    int col[MAXN];
    
    bool dfs(int u)
    {
        for(int i=0;i<G[u].size();i++){
            int v=G[u][i];
            if(!used[v]){
                used[v]=true;
                if(linker[v]==-1||dfs(linker[v])){
                    linker[v]=u;
                    return true;
                }
            }
        }
        return false;
    }
    
    int Hungary()
    {
        int res=0;
        int u;
        memset(linker,-1,sizeof(linker));
        for(u=1;u<=uN;u++){
            memset(used,false,sizeof(used));
            if(dfs(u)) res++;
        }
        return res;
    }
    
    bool Color(int u)
    {
        if(col[u]==-1) col[u]=0;
        for(int i=0;i<G[u].size();i++){
            int v=G[u][i];
            if(col[v]==-1){
                col[v]=col[u]^1;
                if(!Color(v)) return false;
            }
            if(col[v]==col[u]) return false;
        }
        return true;
    }
    
    int main()
    {
        int n,m;
        int u,v;
        while(scanf("%d%d",&n,&m)!=EOF){
            for(int i=1;i<=n;i++) G[i].clear();
            uN=n;
            for(int i=1;i<=m;i++){
                scanf("%d%d",&u,&v);
                G[u].push_back(v);
                G[v].push_back(u);
            }
            bool flag=true;
            memset(col,-1,sizeof(col));
            for(int i=1;i<=n;i++)
                if(col[i]==-1) flag&=Color(i);
            if(!flag){
                printf("No
    ");
                continue;
            }
            else printf("%d
    ",Hungary()/2);
        }
        return 0;
    }
     
  • 相关阅读:
    The Princess and the Pea,摘自iOS应用Snow White and more stories
    Android笔记之自定义的RadioGroup、RadioButton,以及View实例状态的保存与恢复
    Android笔记之获取debug.keystore和release.keystore的MD5/SHA1值
    Android笔记之WebView加载网页的进度回调
    Android Studio停留在“Indexing paused due to batch update”的解决方案
    【Python初级】StringIO和BytesIO读写操作的小思考
    【Python初级】由判定回文数想到的,关于深浅复制,以及字符串反转的问题
    【Python初级】由生成杨辉三角代码所思考的一些问题
    【面试总结-编程】多行两列数据,实现同key的value求和并输出
    【算法与数据结构实战】模拟竖式加法,自定义位数
  • 原文地址:https://www.cnblogs.com/wangdongkai/p/5635723.html
Copyright © 2011-2022 走看看