BZOJ 4761 Cow Navigation

一开始以为瞎jb写个IDA*就过了。。然而并不是这样。(naive)

我的做法是dp[x1][y1][x2][y2][d1][d2]，表示两头奶牛所在的位置和面朝的方向，然后直接spfa搞定。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 25
using namespace std;
int n,map[maxn][maxn],dis[maxn][maxn][maxn][maxn][5][5];
char s[maxn];
int dx[]={0,-1,0,1,0},dy[]={0,0,1,0,-1};
bool vis[maxn][maxn][maxn][maxn][5][5];
struct status
{
    int x1,x2,y1,y2,d1,d2;
    status (int x1,int y1,int x2,int y2,int d1,int d2):x1(x1),y1(y1),x2(x2),y2(y2),d1(d1),d2(d2) {}
    status () {}
};
queue <status> q;
bool judge(int x,int y)
{
    return ((x>=1) && (x<=n) && (y>=1) && (y<=n));
}
void spfa()
{
    q.push(status(n,1,n,1,1,2));
    while (!q.empty())
    {
        status head=q.front();q.pop();
        int nx1,nx2,ny1,ny2;
        nx1=head.x1+dx[head.d1];nx2=head.x2+dx[head.d2];ny1=head.y1+dy[head.d1];ny2=head.y2+dy[head.d2];
        int t2=dis[head.x1][head.y1][head.x2][head.y2][head.d1][head.d2];
        if ((!judge(nx1,ny1)) || (!map[nx1][ny1])) nx1=head.x1,ny1=head.y1;
        if ((!judge(nx2,ny2)) || (!map[nx2][ny2])) nx2=head.x2,ny2=head.y2;
        if ((head.x1==1) && (head.y1==n)) nx1=1,ny1=n;
        if ((head.x2==1) && (head.y2==n)) nx2=1,ny2=n;
        int &t1=dis[nx1][ny1][nx2][ny2][head.d1][head.d2];
        if (t1>t2+1)
        {
            t1=t2+1;
            bool &t3=vis[nx1][ny1][nx2][ny2][head.d1][head.d2];
            if (!t3)
            {
                t3=true;
                q.push(status(nx1,ny1,nx2,ny2,head.d1,head.d2));
            }
        }
        int dd1,dd2;
        dd1=head.d1-1;dd2=head.d2-1;if (!dd1) dd1=4;if (!dd2) dd2=4;
        int &t5=dis[head.x1][head.y1][head.x2][head.y2][dd1][dd2];
        if (t5>t2+1)
        {
            t5=t2+1;
            bool &t3=vis[head.x1][head.y1][head.x2][head.y2][dd1][dd2];
            if (!t3)
            {
                t3=true;
                q.push(status(head.x1,head.y1,head.x2,head.y2,dd1,dd2));
            }
        }
        dd1=head.d1+1;dd2=head.d2+1;if (dd1==5) dd1=1;if (dd2==5) dd2=1;
        int &t4=dis[head.x1][head.y1][head.x2][head.y2][dd1][dd2];
        if (t4>t2+1)
        {
            t4=t2+1;
            bool &t3=vis[head.x1][head.y1][head.x2][head.y2][dd1][dd2];
            if (!t3)
            {
                t3=true;
                q.push(status(head.x1,head.y1,head.x2,head.y2,dd1,dd2));
            }
        }
    }
}
int main()
{
    scanf("%d",&n);memset(dis,0x3f,sizeof(dis));
    for (int i=1;i<=n;i++)
    {
        scanf("%s",s);
        for (int j=0;j<n;j++)
        {
            if (s[j]=='E')
                map[i][j+1]=1;
        }
    }
    dis[n][1][n][1][1][2]=0;
    spfa();
    int mn=0x3f3f3f3f;
    for (int i=1;i<=4;i++)
        for (int j=1;j<=4;j++)
            mn=min(mn,dis[1][n][1][n][i][j]);
    printf("%d
",mn);
    return 0;
}