hdu 1150 Machine Schedule hdu 1151 Air Raid 匈牙利模版

//两道大水……哦不 两道结论题

结论：二部图的最小覆盖数=二部图的最大匹配数

有向图的最小覆盖数=节点数-二部图的最大匹配数

 //hdu 1150
 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<queue>
 #include<vector>
 #include<map>
 #include<stack>
 #include<string>

 using namespace std;

 int n,m,k;
 int f[][];
 int link[];
 bool adj[];
 bool used[];

 bool work(int x){
     for (int i=;i<m;i++){
             if (f[x][i] && adj[i]==false){
                     adj[i]=true;
                     if (!used[i] || work(link[i])){
                             link[i]=x;
                             used[i]=;
                             //printf("%d %d\n",x,i);
                             return true;
                     }
             }
     }
     return false;
 }

 int main(){
     while (scanf("%d%d%d",&n,&m,&k)==){
             memset(f,,sizeof(f));
             memset(used,,sizeof(used));
             for (int i=;i<k;i++){
                     int x,y,z;
                     scanf("%d%d%d",&z,&x,&y);
                     if (x!= && y!=){
                             f[x][y]=;
                     }
             }
             int ans=;
             for (int i=;i<n;i++){
                     memset(adj,,sizeof(adj));
                     if (work(i)) ans++;
             }
             //for (int i=1;i<m;i++) printf("%d %d\n",i,link[i]);
             printf("%d\n",ans);
     }
     return ;
 }
 /*
 5 5 10
 0 1 1
 1 1 2
 2 1 3
 3 1 4
 4 2 1
 5 2 2
 6 2 3
 7 2 4
 8 3 3
 9 4 3
 5 5 10
 0 1 1
 1 1 2
 2 1 3
 3 1 4
 4 2 1
 5 2 2
 6 2 3
 7 2 4
 8 3 3
 9 4 3
 0
 */

 //hdu 1151
 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<queue>
 #include<vector>
 #include<map>
 #include<stack>
 #include<string>

 using namespace std;

 int T;
 int n,m;
 bool f[][];
 int link[];
 bool adj[];
 bool used[];

 bool work(int x){
     for (int i=;i<=n;i++){
             if (f[x][i] && !adj[i]){
                     adj[i]=;
                     if (!used[i] || work(i)){
                             link[i]=x;
                             used[i]=;
                             return true;
                     }
             }
     }
     return false;
 }

 int main(){
     scanf("%d",&T);
     for (int cas=;cas<=T;cas++){
             memset(used,,sizeof(used));
             memset(link,,sizeof(link));
             memset(f,,sizeof(f));
             scanf("%d",&n);
             scanf("%d",&m);
             for (int i=;i<m;i++){
                     int x,y;
                     scanf("%d%d",&x,&y);
                     f[x][y]=;
             }
             int ans=n;
             for (int i=;i<=n;i++){
                     memset(adj,,sizeof(adj));
                     if (work(i)) ans--;
             }
             printf("%d\n",ans);
     }
     return ;
 }
 /*
 2
 4
 3
 3 4
 1 3
 2 3
 3
 3
 1 3
 1 2
 2 3
 */