SPOJ 962 - Intergalactic Map 1<-2->3 构图最大流

题意:

给了一个无向图..要从点1出发..经过点2后达到点3...每个点至多经过一次...问是否存在路径...

题解:

直接搞不好搞..因为是无向图..1->2->3可以看成 1<-2->3...那么就可以很直观的拆点跑网络流了....

Program:

#include<iostream>  
#include<algorithm>  
#include<stdio.h>  
#include<string.h>  
#include<math.h>  
#include<queue>  
#define MAXN 80205
#define MAXM 8000005  
#define oo 1000000007  
#define ll long long  
using namespace std;
struct Dinic        
{        
       struct node      
       {       
             int c,u,v,next;      
       }edge[MAXM];      
       int ne,head[MAXN];      
       int cur[MAXN], ps[MAXN], dep[MAXN];      
       void initial()      
       {      
             ne=2;      
             memset(head,0,sizeof(head));       
       }      
       void addedge(int u, int v,int c)      
       {       
             edge[ne].u=u,edge[ne].v=v,edge[ne].c=c,edge[ne].next=head[u];      
             head[u]=ne++;      
             edge[ne].u=v,edge[ne].v=u,edge[ne].c=0,edge[ne].next=head[v];      
             head[v]=ne++;      
       }      
       int MaxFlow(int s,int t)      
       {                           
             int tr, res = 0;      
             int i,j,k,f,r,top;      
             while(1)      
             {      
                    memset(dep, -1, sizeof(dep));      
                    for(f=dep[ps[0]=s]=0,r=1;f!= r;)      
                       for(i=ps[f++],j=head[i];j;j=edge[j].next)      
                         if(edge[j].c&&dep[k=edge[j].v]==-1)      
                         {      
                               dep[k]=dep[i]+1;      
                               ps[r++]=k;      
                               if(k == t){  f=r; break;  }      
                         }      
                    if(dep[t]==-1) break;      
                    memcpy(cur,head,sizeof(cur));      
                    i=s,top=0;      
                    while(1)      
                    {      
                         if(i==t)      
                         {      
                               for(tr=oo,k=0;k<top;k++)      
                                  if(edge[ps[k]].c<tr)      
                                     tr=edge[ps[f=k]].c;      
                               for(k=0;k<top;k++)      
                               {      
                                     edge[ps[k]].c-=tr;      
                                     edge[ps[k]^1].c+=tr;      
                               }      
                               i=edge[ps[top=f]].u;      
                               res+= tr;      
                         }      
                         for(j=cur[i];cur[i];j=cur[i]=edge[cur[i]].next)       
                             if(edge[j].c && dep[i]+1==dep[edge[j].v]) break;       
                         if(cur[i])  ps[top++]=cur[i],i=edge[cur[i]].v;       
                         else      
                         {      
                                 if(!top) break;      
                                 dep[i]=-1;      
                                 i=edge[ps[--top]].u;      
                         }      
                   }      
             }      
             return res;      
      }      
}T; 
int main() 
{       
      int cases,n,m,s,e,x,y,i;      
      scanf("%d",&cases);
      while (cases--)
      {
              scanf("%d%d",&n,&m);
              s=2<<1,e=(n<<1)+5,T.initial();
              for (i=1;i<=n;i++) T.addedge(i<<1,i<<1|1,1);
              T.addedge(2<<1,2<<1|1,1),T.addedge(3<<1|1,e,1),T.addedge(1<<1|1,e,1);
              while (m--)
              {
                      scanf("%d%d",&x,&y);
                      T.addedge(x<<1|1,y<<1,1),T.addedge(y<<1|1,x<<1,1);
              }
              if (T.MaxFlow(s,e)==2) printf("YES\n");
                               else  printf("NO\n");
      }
      return 0;
}