hdu2063 Hungarian algorithm bipartite graph matching problem Solutions

Topic links: http://acm.hdu.edu.cn/showproblem.php?pid=2063

Hungary augmenting path algorithm is based on a recursive algorithm, applied to the bipartite graph matching problem. This problem is a Hungarian algorithm for maximum matching of bipartite graph application. For V1 a to be matched in the point x, can be considered to the point of him a potential V2 y in a matching object, if this point there is no match over or give y found a new matching objects can be y as an official of x match.

This question I use is the adjacency matrix, each search augmenting path time is O (| V | ^ 2), so the total time complexity is O (| V | ^ 3), the total space complexity is O (| V | ^ 2), if implemented, then the adjacency list total time complexity is O (| V || E |), the spatial complexity is O (| E | + | V |).

code show as below:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef unsigned int ui;
 4 typedef long long ll;
 5 typedef unsigned long long ull;
 6 #define pf printf
 7 #define mem(a,b) memset(a,b,sizeof(a))
 8 #define prime1 1e9+7
 9 #define prime2 1e9+9
10 #define pi 3.14159265
11 #define lson l,mid,rt<<1
12 #define rson mid+1,r,rt<<1|1
13 #define scand(x) scanf("%llf",&x) 
14 #define f(i,a,b) for(int i=a;i<=b;i++)
15 #define scan(a) scanf("%d",&a)
16 #define mp(a,b) make_pair((a),(b))
17 #define P pair<int,int>
18 #define dbg(args) cout<<#args<<":"<<args<<endl;
19 #define inf 0x7ffffff
20 inline int read(){
21     int ans=0,w=1;
22     char ch=getchar();
23     while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}
24     while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
25     return ans*w;
26 }
27 int k,n,m;
28 const int maxn=1e3;
29 int g[maxn][maxn]; 
30 int m_girl,n_boy;
31 int match[maxn],reserve_boy[maxn];//匹配结果在match中 
32 boolDFS ( int x) // Hungarian algorithm: for x to find an augmenting path, i.e. to find a matching object 
33 is  {
 34 is      for ( int I = . 1 ; I <= n_boy; I ++ )
 35      {
 36          IF (reserve_boy [I! ] && G [x] [i]) // boy i and girls are not preserve object x may be matched and i 
37 [          {
 38 is              reserve_boy [i] = . 1 ; // predetermined boy i, ready to give x 
39              IF (match! [i] || DFS (match [i]))
 40              // there are two cases, (1) the boy i no matching object (2) boys paired objects have now attempt to replace the boy i Original objects both a successful i assigned to put X 
41 is              {
 42 is                 match [i] = X; // i-th pair boy or directly to the object to replace X 
43 is                  return  to true ;
 44 is              } 
 45           } 
 46 is      }
 47      return  to false ; // girls i pairing fails 
48  }
 49  int main ()
 50  {
 51 is      // The freopen ( "input.txt", "R & lt", stdin);
 52 is      // The freopen ( "output.txt", "W", stdout); 
53 is      STD :: :: iOS sync_with_stdio ( to false );
 54 is      the while (Scanf ( " % D " , & K) ==1&&k)
55     {
56         m_girl=read(),n_boy=read();
57         mem(g,0);
58         mem(match,0); 
59         int u,v;
60         f(i,1,k)
61         {
62             u=read(),v=read();
63             g[u][v]=1;//邻接矩阵存边 
64         }
65         int sum=0;
66         f(i,1,m_girl)//Looking for each girl pair, the current seek successful girl after may be replaced, but will successfully paired 
67          {
 68              MEM (reserve_boy, 0 );
 69              IF (DFS (I)) SUM ++ ;
 70          }
 71 is          PF ( " % D \ n- " , SUM);
 72      }
 73 is }