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;}