ZOJ 2760 - How Many Shortest Path(最短路的个数) Floyd+最大流

题意:

给一个图..问s到e的最短距离有多少个...

题解:

之前做过一个相似的..用最小费用最大流来做..也就是限制住每次的费用为最短距离..跑到不能跑..但是WA了...

参考小媛的..用Floyd+最大流解决...挺给力的...

Program:

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#define MAXN 1005
#define MAXM 100005
#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 A[105][105],B[105][105]; 
void Floyd(int n)
{
      int k,i,j;
      for (k=0;k<n;k++)
         for (i=0;i<n;i++)
            for (j=0;j<n;j++)
            { 
                    if (A[i][k]<0 || A[k][j]<0) continue;
                    if (A[i][j]!=-1 && A[i][j]<A[i][k]+A[k][j]) continue;
                    A[i][j]=A[i][k]+A[k][j];
            }
}
int main() 
{             
      int n,i,j,s,e;   
      while (~scanf("%d",&n))
      {
               for (i=0;i<n;i++)
                  for (j=0;j<n;j++)
                  {
                       scanf("%d",&A[i][j]);
                       if (i==j) A[i][j]=0; 
                  }
               memcpy(B,A,sizeof(B));
               Floyd(n);
               scanf("%d%d",&s,&e);
               if (s==e)
               {
                       printf("inf\n");
                       continue;
               } 
               T.initial(); 
               for (i=0;i<n;i++)
                  if (A[s][i]!=-1)
                    for (j=0;j<n;j++)
                      if (B[i][j]!=-1 && A[s][j]!=-1 &&  A[s][i]+B[i][j]==A[s][j])
                       T.addedge(i,j,1);
               printf("%d\n",T.MaxFlow(s,e));
      }
      return 0;
}