BZOJ3144: [Hnoi2013]Cut cake

【Portal: BZOJ3144】

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Brief title:

　　

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answer:

　　Minimum cut classical problem

　　We construct a lattice of P*Q*(R+1) and use (i,j,k) to represent a point, then the flow of (i,j,k)->(i,j,k+1) is the original The dissonance value of (i,j,k) in the graph. S connects infinite edges to all bottom layers, and all top layers connect infinite edges to T. Then (i,j,k) -> (i,j,k+1) is cut to indicate f(i,j)=k

　　At the same time, we let (i,j,k)->(i',j',kD) connect an infinite edge, and we can ensure that the two adjacent ones do not exceed D.

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Reference Code:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define INF 1<<31-1
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct node
{
    int x,y,c,next,other;
}a[1100000];int len,last[110000];
void ins(int x,int y,int c)
{
    int k1 = ++ len, k2 = ++ len;
    a[k1].x =x;a[k1].y=y;a[k1].c= c;
    a[k1].next=last[x];last[x]=k1;
    a[k2].x=y;a[k2].y=x;a[k2].c=0;
    a[k2].next=last[y];last[y]=k2;
    a[k1].other=k2;
    a[k2].other=k1;
}
int h[110000],list[110000],st,ed;
bool bt_h()
{
    memset(h,0,sizeof(h));h[st]=1;
    list[1]=st;
    int head=1,tail=2;
    while(head!=tail)
    {
        int x=list[head];
        for(int k=last[x];k;k=a[k].next)
        {
            int y=a[k].y;
            if(a[k].c>0&&h[y]==0)
            {
                h[y]=h[x]+1;
                list[tail++]=y;
            }
        }
        head++;
    }
    if(h[ed]==0) return false;
    else return true;
}
int cur[110000];
int findflow(int x,int f)
{
    if(x==ed) return f;
    int s=0,t;
    for(int k=cur[x];k;k=a[k].next)
    {
        int y=a[k].y;
        if(a[k].c>0&&f>s&&h[y]==(h[x]+1))
        {
            t=findflow(y,min(a[k].c,f-s));
            s+=t;
            a[k].c-=t;a[a[k].other].c+=t;
            if(a[k].c>0) cur[x]=k;
            if(f==s) break;
        }
    }
    if(s==0) h[x]=0;
    return s;
}
int v[41][41][41],id[41][41][41];
int dx[5]={0,1,-1,0,0};
int dy[5]={0,0,0,1,-1};
int main()
{
    int P=read(),Q=read(),R=read(),D=read();
    int z=0;
    for(int i=1;i<=P;i++) for(int j=1;j<=Q;j++) id[0][i][j]=++z;
    for(int i=1;i<=R;i++) for(int j=1;j<=P;j++) for(int k=1;k<=Q;k++) v[i][j][k]=read(),id[i][j][k]=++z;
    len=0;memset(last,0,sizeof(last));
    st=0;ed=z+1;
    for(int i=0;i<=R;i++)
    {
        if(i==R) for(int j=1;j<=P;j++) for(int k=1;k<=Q;k++) ins(id[i][j][k],ed,INF);
        if(i==0) for(int j=1;j<=P;j++) for(int k=1;k<=Q;k++) ins(st,id[i][j][k],INF);
        else for(int j=1;j<=P;j++) for(int k=1;k<=Q;k++) ins(id[i-1][j][k],id[i][j][k],v[i][j][k]);
    }
    for(int i=D+1;i<=R;i++)
    {
        for(int j=1;j<=P;j++)
        {
            for(int k=1;k<=Q;k++)
            {
                for(int t=1;t<=4;t++)
                {
                    int tx=j+dx[t],ty=k+dy[t];
                    if(tx<1||ty<1||tx>P||ty>Q) continue;
                    ins(id[i][j][k],id[i-D][tx][ty],999999999);
                }
            }
        }
    }
    int ans=0;
    while(bt_h()==true)
    {
        for(int i=st;i<=ed;i++) cur[i]=last[i];
        ans += findflow(st,INF);
    }
    printf("%d\n",ans);
    return 0;
}