ICPC-Beijing 2006 狼抓兔子

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1100;
template<typename T>
inline void read(T&x)
{
    T f = 1,c = 0;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){c=c*10+ch-‘0‘;ch=getchar();}
    x = f*c;
}
int n,m,s,t,hed[2*N*N],cnt;
int _id(int i, int j, int k) {
    if(i>n||j<1)return 0;
    if(i<1||j>m)return n*m<<1|1;
    return (i - 1) * m + j + k * n * m;
}
struct EG
{
    int to,nxt;
    ll w;
}e[6*N*N];
void ae(int f,int t,ll w)
{
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    e[cnt].w = w;
    hed[f] = cnt;
}
struct Pair
{
    int x;
    ll d;
    Pair(){}
    Pair(int x,ll d):x(x),d(d){}
    friend bool operator < (Pair a,Pair b){return a.d>b.d;}
}tp;
ll dis[2*N*N];
bool vis[2*N*N];
void dij()
{
    priority_queue<Pair>q;
    memset(dis,0x3f,sizeof(dis));dis[s]=0;
    q.push(Pair(s,0));
    while(!q.empty())
    {
        tp = q.top();
        q.pop();
        int u = tp.x;
        if(vis[u])continue;
        vis[u]=1;
        for(int j=hed[u];j;j=e[j].nxt)
        {
            int to = e[j].to;
            if(dis[to]>dis[u]+e[j].w)
            {
                dis[to] = dis[u]+e[j].w;
                q.push(Pair(to,dis[to]));
            }
        }
    }
}
int main()
{
    read(n),read(m);
    n--,m--;
    s = 0,t = n*m<<1|1;
    for(int i=1;i<=n+1;i++)
        for(int w,j=1;j<=m;j++)
        {
            read(w);
            ae(_id(i,j,1),_id(i-1,j,0),w);
            ae(_id(i-1,j,0),_id(i,j,1),w);
        }
    for(int i=1;i<=n;i++)
        for(int w,j=1;j<=m+1;j++)
        {
            read(w);
            ae(_id(i,j,0),_id(i,j-1,1),w);
            ae(_id(i,j-1,1),_id(i,j,0),w);
        }
    for(int i=1;i<=n;i++)
        for(int w,j=1;j<=m;j++)
        {
            read(w);
            ae(_id(i,j,0),_id(i,j,1),w);
            ae(_id(i,j,1),_id(i,j,0),w);
        }
    dij();
    printf("%lld\n",dis[t]);
    return 0;
}