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POJ 3041 Asteroids 二分图匹配

时间:2014-05-08 14:32:28      阅读:347      评论:0      收藏:0      [点我收藏+]

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

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//#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!=\n)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\n",solve());
     }
     return 0;
}
View Code
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POJ 3041 Asteroids 二分图匹配,布布扣,bubuko.com

POJ 3041 Asteroids 二分图匹配

原文:http://www.cnblogs.com/BMan/p/3713655.html

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