[状压dp][spfa] Jzoj P3737 挖宝藏

Description

 

 

题解

• 我们可以发现宝藏数最多只会有9个，显然可以状压
• 而且数据这么小，就可以愉快
• 考虑dp，设f[i][j][k][s]表示在第i层(i,j)宝藏集合的状态为s
• 状态转移方程就是f[i][j][k][s]=min(f[i][j][k][s1]+f[i][j][s-s1]-a[x][y])(s包含s1)
• 还有向四个方向拓展，可以用spfa松弛
• 对于换层操作，我们在所到层的同一位置f[i-1][x][y][1]=f[i][x][y][mi[num[i]+1]-1]

 

代码

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int N=15,M=1<<12,dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int h,n,m,l,r,ans=1e8,mi[M],a[N][N][N],f[N][N][N][M],num[N],d[N*N][3],b[N][N][3],bz[N][N];
inline int read()
{
    int s=0,w=1;
       char ch=getchar();
       while (ch<'0'||ch>'9'){ if(ch=='-') w=-1;ch=getchar(); }
       while (ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
       return s*w;
}
int main()
{
    freopen("treasure.in","r",stdin),freopen("treasure.out","w",stdout);
    freopen("data.in","r",stdin);
    h=read(),n=read(),m=read();
    mi[0]=1; for (register int i=1;i<=11;++i) mi[i]=mi[i-1]*2;
    for (register int i=1;i<=h;++i) for (register int j=1;j<=n;++j) for (register int k=1;k<=m;k++)
    {
        a[i][j][k]=read();
        for (register int l=0;l<mi[11];l++) f[i][j][k][l]=1e8;
    }
    for (register int i=1;i<=h;++i)
    {
        num[i]=read();
        for (register int j=1;j<=num[i];++j) b[i][j][0]=read(),b[i][j][1]=read();
    }
    for (register int i=h;i>=1;i--)
    {
        for (register int x=1;x<=n;++x) for (register int y=1;y<=m;++y) f[i][x][y][mi[num[i]]]=f[i+1][x][y][mi[num[i+1]+1]-1]+a[i][x][y],f[i][x][y][0]=a[i][x][y];
        for (register int j=1;j<=num[i];++j) f[i][b[i][j][0]][b[i][j][1]][mi[j-1]]=a[i][b[i][j][0]][b[i][j][1]];
        for (register int s=1;s<mi[num[i]+1];s++)
            for (register int x=1;x<=n;++x) for (register int y=1;y<=m;++y)
            {
                for (register int s1=1;s1<=s;s1++) if ((s&s1)==s1) if (f[i][x][y][s1]+f[i][x][y][s-s1]-a[i][x][y]<f[i][x][y][s]) f[i][x][y][s]=f[i][x][y][s1]+f[i][x][y][s-s1]-a[i][x][y];
                memset(bz,0,sizeof(bz)),d[1][0]=x,d[1][1]=y,l=0,r=1;
                while (l<r)
                {
                    l++;
                    for (register int k=0;k<4;k++)
                    {
                        register int xx=d[l][0]+dx[k],yy=d[l][1]+dy[k];
                        if (xx>0&&xx<=n&&yy>0&&yy<=m&&f[i][xx][yy][s]>f[i][d[l][0]][d[l][1]][s]+a[i][xx][yy])
                        {
                            f[i][xx][yy][s]=f[i][d[l][0]][d[l][1]][s]+a[i][xx][yy];
                            if (!bz[xx][yy]) bz[xx][yy]=1,d[++r][0]=xx,d[r][1]=yy;
                        }
                    }
                    bz[d[l][0]][d[l][1]]=0;
                }
            }
    }
    for (register int i=1;i<=n;++i) for (register int j=1;j<=m;++j) ans=min(ans,f[1][i][j][mi[num[1]+1]-1]);
    printf("%d",ans);
}