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  • 网络流输出二分匹配方案

    题意:https://www.luogu.org/problem/P2756

      1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
      2 #include <cstdio>//sprintf islower isupper
      3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
      4 #include <iostream>//pair
      5 #include <fstream>//freopen("C:\Users\13606\Desktop\草稿.txt","r",stdin);
      6 #include <bitset>
      7 //#include <map>
      8 //#include<unordered_map>
      9 #include <vector>
     10 #include <stack>
     11 #include <set>
     12 #include <string.h>//strstr substr
     13 #include <string>
     14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
     15 #include <cmath>
     16 #include <deque>
     17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
     18 #include <vector>//emplace_back
     19 //#include <math.h>
     20 #include <cassert>
     21 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
     22 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
     23 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
     24 //******************
     25 int abss(int a);
     26 int lowbit(int n);
     27 int Del_bit_1(int n);
     28 int maxx(int a,int b);
     29 int minn(int a,int b);
     30 double fabss(double a);
     31 void swapp(int &a,int &b);
     32 clock_t __STRAT,__END;
     33 double __TOTALTIME;
     34 void _MS(){__STRAT=clock();}
     35 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
     36 //***********************
     37 #define rint register int
     38 #define fo(a,b,c) for(rint a=b;a<=c;++a)
     39 #define fr(a,b,c) for(rint a=b;a>=c;--a)
     40 #define mem(a,b) memset(a,b,sizeof(a))
     41 #define pr printf
     42 #define sc scanf
     43 #define ls rt<<1
     44 #define rs rt<<1|1
     45 typedef vector<int> VI;
     46 typedef long long ll;
     47 const double E=2.718281828;
     48 const double PI=acos(-1.0);
     49 //const ll INF=(1LL<<60);
     50 const int inf=(1<<30);
     51 const double ESP=1e-9;
     52 const int mod=(int)1e9+7;
     53 const int N=(int)2e5+10;
     54 
     55 class DINIC
     56 {
     57 public:
     58 //    const int MAXN=10004,MAXWAY=100005;
     59     int n,way,max_flow,deep[N];
     60     int tot,head[N],cur[N];
     61     struct EDGE{
     62         int to,next;
     63         int dis;
     64     }edge[N];
     65     void Init(int n)
     66     {
     67         tot=-1;//因为加反向边要^1,所以要从0开始;
     68         this->n=n;
     69         for(int i=0;i<=n;++i)
     70             head[i]=-1;
     71     }
     72     void add(int from,int to,int V,bool flag)
     73     {
     74         ++tot;
     75         edge[tot].to=to;
     76         edge[tot].dis=0;
     77         if(flag)edge[tot].dis=V;
     78         edge[tot].next=head[from];
     79         head[from]=tot;
     80     }
     81     queue<int>q;
     82     bool bfs(int s,int t)
     83     {
     84         for(int i=1;i<=n;++i)
     85             deep[i]=inf;
     86         while(!q.empty())q.pop();
     87         for(int i=1;i<=n;++i)cur[i]=head[i];
     88         deep[s]=0;
     89         q.push(s);
     90 
     91         while(!q.empty())
     92         {
     93             int now=q.front();q.pop();
     94             for(int i=head[now];i!=-1;i=edge[i].next)
     95             {
     96                 if(deep[edge[i].to]==inf&&edge[i].dis)
     97                 {
     98                     deep[edge[i].to]=deep[now]+1;
     99                     q.push(edge[i].to);
    100                 }
    101             }
    102         }
    103         return deep[t]<inf;
    104     }
    105     int dfs(int now,int t,int limit)
    106     {
    107         if(!limit||now==t)return limit;
    108         int flow=0,f;
    109         for(int i=cur[now];i!=-1;i=edge[i].next)
    110         {
    111             cur[now]=i;
    112             if(deep[edge[i].to]==deep[now]+1&&(f=dfs(edge[i].to,t,min(limit,edge[i].dis))))
    113             {
    114                 flow+=f;
    115                 limit-=f;
    116                 edge[i].dis-=f;
    117                 edge[i^1].dis+=f;
    118                 if(!limit)break;
    119             }
    120         }
    121         return flow;
    122     }
    123     void Dinic(int s,int t)
    124     {
    125         while(bfs(s,t))
    126             max_flow+=dfs(s,t,inf);
    127     }
    128 }G;
    129 
    130 int main()
    131 {
    132     int m,n;
    133     sc("%d%d",&m,&n);
    134     int u,v,cnt=0;
    135     G.Init(n+2);
    136     while(sc("%d%d",&u,&v),u!=-1&&v!=-1)
    137     {
    138         u++,v++;
    139         cnt++;
    140         G.add(u,v,1,1);G.add(v,u,1,0);
    141     }
    142     cnt*=3;
    143     for(int i=2;i<=m+1;++i)
    144     {
    145         G.add(1,i,1,1);
    146         G.add(i,1,1,0);
    147     }
    148     for(int i=m+2;i<=n+1;++i)
    149     {
    150         G.add(i,n+2,1,1);
    151         G.add(n+2,i,1,0);
    152     }
    153     G.Dinic(1,n+2);
    154     pr("%d
    ",G.max_flow);
    155     for(int i=1,j=0;i<=cnt;++i,j+=2)
    156     {
    157         if(G.edge[j].to!=1&&G.edge[j+1].to!=1)
    158             if(G.edge[j].to!=n+2&&G.edge[j+1].to!=n+2)
    159                 if(G.edge[j+1].dis)//网络流可以判断正向边是否有流量(即反向边的残量是否不为 0)
    160                 {
    161                     pr("%d %d
    ",G.edge[j+1].to-1,G.edge[j].to-1);
    162                 }
    163     }
    164     return 0;
    165 }
    166 
    167 /**************************************************************************************/
    168 
    169 int maxx(int a,int b)
    170 {
    171     return a>b?a:b;
    172 }
    173 
    174 void swapp(int &a,int &b)
    175 {
    176     a^=b^=a^=b;
    177 }
    178 
    179 int lowbit(int n)
    180 {
    181     return n&(-n);
    182 }
    183 
    184 int Del_bit_1(int n)
    185 {
    186     return n&(n-1);
    187 }
    188 
    189 int abss(int a)
    190 {
    191     return a>0?a:-a;
    192 }
    193 
    194 double fabss(double a)
    195 {
    196     return a>0?a:-a;
    197 }
    198 
    199 int minn(int a,int b)
    200 {
    201     return a<b?a:b;
    202 }
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  • 原文地址:https://www.cnblogs.com/--HPY-7m/p/11755260.html
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