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  • POJ 3041 Asteroids 二分图匹配

    以行列为点建图,每个点(x,y) 对应一条边连接x,y。二分图的最小点覆盖=最大匹配

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<iostream>
    #include<sstream>
    #include<cmath>
    #include<climits>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    #define pb(a) push(a)
    #define INF 0x1f1f1f1f
    #define lson idx<<1,l,mid
    #define rson idx<<1|1,mid+1,r
    #define PI  3.1415926535898
    template<class T> T min(const T& a,const T& b,const T& c) {
        return min(min(a,b),min(a,c));
    }
    template<class T> T max(const T& a,const T& b,const T& c) {
        return max(max(a,b),max(a,c));
    }
    void debug() {
    #ifdef ONLINE_JUDGE
    #else
    
        freopen("data.in","r",stdin);
       // freopen("d:\out1.txt","w",stdout);
    #endif
    }
    int getch() {
        int ch;
        while((ch=getchar())!=EOF) {
            if(ch!=' '&&ch!='
    ')return ch;
        }
        return EOF;
    }
    
    const int maxn = 1100;
    vector<int> g[maxn];
    int n, m;
    void Init()
    {
         for(int i=1; i<=n; i++)
              g[i].clear();
    }
    
    void add(int u, int v)
    {
         g[u].push_back(v);
    }
    
    int match[maxn];
    int vis[maxn];
    
    bool dfs(int u)
    {
         vis[u] = true;
         for(int i = 0; i < g[u].size(); i++)
         {
              int v = g[u][i];
              int w = match[v];
              if(w<0||!vis[w]&&dfs(w))
              {
                   match[u] = v;
                   match[v] = u;
                   return true;
              }
         }
         return false;
    }
    int solve()
    {
         memset(match,-1,sizeof(match));
         int res = 0;
         for(int u=1; u <= n; u++)
         {
              if(match[u]<0)
              {
                   memset(vis, 0, sizeof(vis));
                   if(dfs(u))
                        res++;
              }
         }
         return res;
    }
    int main()
    {
         debug();
         while(scanf("%d%d", &n, &m) != EOF)
         {
              for(int i=1; i<=m; i++)
              {
                   int u,v;
                   scanf("%d%d", &u, &v);
                   add(u, v+n);
                   add(v+n, u);
              }
    
              printf("%d
    ",solve());
         }
         return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/BMan/p/3713655.html
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