POJ 1815 Friendship（最小割+字典序输出割点）

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<sstream>
  6 #include<vector>
  7 #include<stack>
  8 #include<queue>
  9 #include<cmath>
 10 #include<map>
 11 #include<set>
 12 using namespace std;
 13 typedef long long ll;
 14 typedef pair<int,int> pll;
 15 const int INF = 0x3f3f3f3f;
 16 const int maxn = 20000 + 5;
 17 
 18 struct Edge
 19 {
 20     int from,to,cap,flow;
 21     Edge(int u,int v,int w,int f):from(u),to(v),cap(w),flow(f){}
 22 };
 23 
 24 struct Dinic
 25 {
 26     int n,m,s,t;
 27     vector<Edge> edges;
 28     vector<int> G[maxn];
 29     bool vis[maxn];
 30     int cur[maxn];
 31     int d[maxn];
 32 
 33     void init(int n)
 34     {
 35         this->n=n;
 36         for(int i=0;i<n;++i) G[i].clear();
 37         edges.clear();
 38     }
 39 
 40     void AddEdge(int from,int to,int cap)
 41     {
 42         edges.push_back( Edge(from,to,cap,0) );
 43         edges.push_back( Edge(to,from,0,0) );
 44         m=edges.size();
 45         G[from].push_back(m-2);
 46         G[to].push_back(m-1);
 47     }
 48 
 49     bool BFS()
 50     {
 51         queue<int> Q;
 52         memset(vis,0,sizeof(vis));
 53         vis[s]=true;
 54         d[s]=0;
 55         Q.push(s);
 56         while(!Q.empty())
 57         {
 58             int x=Q.front(); Q.pop();
 59             for(int i=0;i<G[x].size();++i)
 60             {
 61                 Edge& e=edges[G[x][i]];
 62                 if(!vis[e.to] && e.cap>e.flow)
 63                 {
 64                     vis[e.to]=true;
 65                     d[e.to]=d[x]+1;
 66                     Q.push(e.to);
 67                 }
 68             }
 69         }
 70         return vis[t];
 71     }
 72 
 73     int DFS(int x,int a)
 74     {
 75         if(x==t || a==0) return a;
 76         int flow=0, f;
 77         for(int &i=cur[x];i<G[x].size();++i)
 78         {
 79             Edge &e=edges[G[x][i]];
 80             if(d[e.to]==d[x]+1 && (f=DFS(e.to,min(a,e.cap-e.flow) ) )>0)
 81             {
 82                 e.flow +=f;
 83                 edges[G[x][i]^1].flow -=f;
 84                 flow +=f;
 85                 a -=f;
 86                 if(a==0) break;
 87             }
 88         }
 89         return flow;
 90     }
 91 
 92     int Maxflow(int s,int t)
 93     {
 94         this->s=s; this->t=t;
 95         int flow=0;
 96         while(BFS())
 97         {
 98             memset(cur,0,sizeof(cur));
 99             flow +=DFS(s,INF);
100         }
101         return flow;
102     }
103 }DC;
104 
105 int n,s,t;
106 int mp[maxn][maxn];
107 
108 int main()
109 {
110     //freopen("in.txt","r",stdin);
111     while(~scanf("%d%d%d",&n,&s,&t))
112     {
113         int dst=2*n+1;
114         DC.init(dst+1);
115 
116         for(int i=1;i<=n;i++)
117         {
118             if(i==s || i==t)   DC.AddEdge(i,i+n,INF);
119             else DC.AddEdge(i,i+n,1);
120         }
121 
122         for(int i=1;i<=n;i++)
123         {
124             for(int j=1;j<=n;j++)
125             {
126                 scanf("%d",&mp[i][j]);
127                 if(i==j)  continue;
128                 if(mp[i][j]) DC.AddEdge(i+n,j,INF);
129             }
130         }
131 
132         if(mp[s][t])  {puts("NO ANSWER!");continue;}
133 
134         int cnt=DC.Maxflow(s,t);
135         vector<int> ans;
136 
137         printf("%d\n",cnt);
138         if(cnt==0)   continue;
139 
140         for(int i=0;i<2*n;i+=2)  //前2*n条边为每个顶点，依次枚举
141         {
142             Edge& e=DC.edges[i];
143             if(e.from==s || e.from==t)  continue;
144             e.cap=0;  //删边
145             for(int j=0;j<2*n;j++) DC.edges[j].flow = 0;
146             int tmp=DC.Maxflow(s,t);  //如果流量改变，说明这条边是割边
147             if(tmp<cnt)
148             {
149                 cnt=tmp;
150                 ans.push_back(e.from);
151             }
152             else e.cap=1;
153             if(cnt<=0)  break;
154         }
155         for(int i=0;i<ans.size();i++)
156         {
157             printf("%d%c",ans[i],i==ans.size()-1?‘\n‘:‘ ‘);
158         }
159         printf("\n");
160     }
161     return 0;
162 }