UVA 1078 Steam Roller 建图 Dijksta

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;
}