Poj 3271 Lilypad Pond bfs

Because of the existence of 1, the problem becomes more difficult, so it is relatively simple to remove 1, first bfs, process the points that each point can reach in one step (it must be one step ). then, we can use bfs on the New Graph to calculate the number of the shortest paths and the shortest paths between the two points. (As to why this is the original question, it is very simple. Because the construction of the xiangpu should be the least, it will not be repeated, and there will be no more construction, so it is to find the shortest path. As for the number of paths, because the path length is simply incrementing, you can simply accumulate it ).

 #include <iostream>  #include <cstdio>  #include <cstring>  using namespace std;  const int maxn=30+9;  int dist[maxn][maxn],a[maxn][maxn];  int n,m;  bool d[maxn][maxn][maxn][maxn];  int quex[maxn*maxn],quey[maxn*maxn];  long long ans[maxn][maxn];  bool text[maxn][maxn];  void bfs2(int t,int s)  {      int front=1,end=0;      quex[++end]=t;      quey[end]=s;      memset(dist,50,sizeof(dist));      memset(ans,0,sizeof(ans));      dist[t][s]=0;      ans[t][s]=1;      while(front<=end)      {          int x=quex[front],y=quey[front++];          if(a[x][y]==2) continue;          for(int i=1;i<=n;i++)          for(int j=1;j<=m;j++)          if(d[x][y][i][j])          {              if(dist[i][j]==dist[x][y]+1)              ans[i][j]+=ans[x][y];              else if(dist[i][j]>dist[x][y]+1)              {                  dist[i][j]=dist[x][y]+1;                  ans[i][j]=ans[x][y];                  quex[++end]=i;                  quey[end]=j;              }          }      }  }    void bfs(int t,int s)  {      memset(text,0,sizeof(text));      int front=1,end=0;      quex[++end]=t;      quey[end]=s;      text[t][s]=1;      while(front<=end)      {          int x=quex[front],y=quey[front++];          for(int i=-1;i<=1;i+=2)          for(int j=-1;j<=1;j+=2)          for(int k=1;k<=2;k++)          {              int tmp=1;              if(k==1) tmp=2;              int tox=x+i*k,toy=j*tmp+y;              if(tox>=1&&tox<=n&&toy>=1&&toy<=m)              if(!text[tox][toy])              {                  text[tox][toy]=1;                  if(a[tox][toy]==1)                  {                      quex[++end]=tox;                      quey[end]=toy;                  }                  else                  {                      d[t][s][tox][toy]=1;                  }              }          }      }  }    int main()  {      while(scanf("%d %d",&n,&m)!=EOF)      {          int t,s,tox,toy;          for(int i=1;i<=n;i++)          for(int j=1;j<=m;j++)          {              scanf("%d",&a[i][j]);              if(a[i][j]==3)              {                  t=i;                  s=j;              }              else if(a[i][j]==4)              {                  tox=i;                  toy=j;              }          }          memset(d,0,sizeof(d));          for(int i=1;i<=n;i++)          for(int j=1;j<=m;j++)          bfs(i,j);          bfs2(t,s);          if(dist[tox][toy]>1e3)          {              printf("-1\n");          }          else          {              printf("%d\n%lld\n",dist[tox][toy]-1,ans[tox][toy]);          }      }      return 0;  }