[洛谷P3386][题解][模板]二分图匹配

题目戳我

利用匈牙利算法实现二分图最大匹配

主要是递归和贪心的思想

Code:

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 #define rg register
 4 #define us unsigned
 5 #define eps 1e-6
 6 #define INF 0x3f3f3f3f
 7 #define ls k<<1
 8 #define rs k<<1|1
 9 #define tmid (tr[k].l+tr[k].r)>>1
10 #define nmid (l+r)>>1
11 #define Thispoint tr[k].l==tr[k].r
12 #define pushup tr[k].wei=tr[ls].wei+tr[rs].wei
13 using namespace std;
14 inline void Read(int &x){
15     int f=1;
16     char c=getchar();
17     x=0;
18     while(c<'0'||c>'9'){
19         if(c=='-')f=-1;
20         c=getchar();
21     }
22     while(c>='0'&&c<='9'){
23         x=(x<<3)+(x<<1)+c-'0';
24         c=getchar();
25     }
26     x*=f;
27 }
28 inline void Print(int x){
29     if(x<0){
30         x=~(x-1);
31         putchar('-');
32     }
33     if(x>9)Print(x/10);
34     putchar(x%10+'0');
35 }
36 #define N 1010
37 int n,m,e,ans,match[N];
38 bool mp[N][N],vis[N];
39 bool DFS(int x){
40     for(int i=1;i<=m;i++){
41         if(!vis[i]&&mp[x][i]){
42             vis[i]=1;
43             if(!match[i]||DFS(match[i])){
44                 match[i]=x;
45                 return 1;
46             }
47         }
48     }
49     return 0;
50 }
51 int main(){
52     Read(n),Read(m),Read(e);
53     for(int i=1;i<=e;i++){
54         int u,v;
55         Read(u),Read(v);
56         if(v<=m&&u<=n)mp[u][v]=1;
57     }
58     for(int i=1;i<=n;i++){
59         ans+=DFS(i);
60         memset(vis,0,sizeof(vis));
61     }
62     Print(ans);
63     return 0;
64 }

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转载自www.cnblogs.com/juruoajh/p/12303393.html