把一个点(r,c)拆成（r,c,dir,doubled)八个点
表示上个点是从dir方向到(r,c)的，doubled表示那条边是否已经加倍。
而后就是考虑清楚细节，建图。
最后跑最短路。
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("d:\in1.txt","r",stdin); freopen("d:\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!=' ')return ch; } return EOF; } struct HeapNode { int d,u; bool operator < (const HeapNode &ant) const { return d>ant.d; } }; struct Edge { int from,to; int dist; }; const int maxc=105; const int maxr=105; const int maxn=8*maxr*maxc; struct Dijksta { int n; vector<int> g[maxn]; vector<Edge> edge; int d[maxn]; int done[maxn]; void init(int n) { this->n = n; for(int i=0;i<n;i++) g[i].clear(); edge.clear(); } void add(int u,int v,int w) { Edge e=(Edge){u,v,w}; edge.push_back(e); g[u].push_back(edge.size()-1); } void solve(int s) { memset(d,INF,sizeof(d)); memset(done,0,sizeof(done)); d[s]=0; priority_queue<HeapNode> q; q.push((HeapNode){0,s}); while(!q.empty()) { HeapNode x=q.top();q.pop(); if(done[x.u])continue; int u=x.u; done[u]=1; for(int i=0;i<g[u].size();i++) { Edge &e=edge[g[u][i]]; if(e.dist+d[u]<d[e.to]) { d[e.to]=d[u]+e.dist; q.push((HeapNode){d[e.to],e.to}); } } } } int findD(int e) { return d[e]; } }; int R,C,r1,r2,c1,c2; int n; int grid[maxr][maxc][4]; const int UP=0,RIGHT=1,DOWN=2,LEFT=3; int inv[]={2,3,0,1}; int dr[]={-1,0,1,0}; int dc[]={0,1,0,-1}; Dijksta solver; int id[maxr][maxc][4][2]; int ID(int r,int c,int dir,int doubled) { int &x=id[r][c][dir][doubled]; if(x>0)return x; else return x=++n; } int readint() { int x;scanf("%d",&x);return x; } void read() { for(int r=1;r<=R;r++) { for(int c=1;c<C;c++) grid[r][c][RIGHT]=grid[r][c+1][LEFT]=readint(); if(r!=R)for(int c=1;c<=C;c++) grid[r][c][DOWN]=grid[r+1][c][UP]=readint(); } } bool cango(int r,int c,int dir) { if(r<1||r>R||c<1||c>C)return false; return grid[r][c][dir]>0; } void construct() { for(int r=1;r<=R;r++) for(int c=1;c<=C;c++) for(int d=0;d<4;d++)if(cango(r,c,inv[d])) for(int nd=0;nd<4;nd++)if(cango(r,c,nd)) for(int doubled=0;doubled<2;doubled++) { int nr=r+dr[nd]; int nc=c+dc[nd]; int v=grid[r][c][nd]; int ndoubled=0; if(d!=nd) { if(!doubled)v+=grid[r][c][inv[d]]; ndoubled=1;v+=grid[r][c][nd]; } solver.add(ID(r,c,d,doubled),ID(nr,nc,nd,ndoubled),v); } } int main() { int ca=0; while(scanf("%d%d%d%d%d%d",&R,&C,&r1,&c1,&r2,&c2)!=EOF&&R) { read(); n=0; memset(id,0,sizeof(id)); solver.init(R*C*8+1); for(int d=0;d<4;d++)if(cango(r1,c1,d)) solver.add(0,ID(r1+dr[d],c1+dc[d],d,1),grid[r1][c1][d]*2); construct(); solver.solve(0); int ans=INF; for(int d=0;d<4;d++)if(cango(r2,c2,inv[d])) for(int doubled=0;doubled<2;doubled++) { int v=solver.findD(ID(r2,c2,d,doubled)); if(!doubled)v+=grid[r2][c2][inv[d]]; ans=min(ans,v); } printf("Case %d: ",++ca); if(ans!=INF)printf("%d ",ans); else printf("Impossible "); } return 0; }