BZOJ3144: [Hnoi2013] cut cake min cut

Source: Internet
Author: User
Description

Input

The first line is a three positive integer p,q,r, representing the cut cake's long p, wide Q, and high R. The second row has a nonnegative integer d, which indicates smoothness requirements. Next is the matrix of R P row Q columns, the x row of the z matrix is V (x,y,z) (1≤x≤p, 1≤y≤q, 1≤z≤r).
100% of the data satisfies the p,q,r≤40,0≤d≤r, and all the disharmony values given are not more than 1000. Output

Contains only an integer that represents the smallest total disharmony value on a legal basis. Sample Input

2 2 2

1

6 1

6 1

2 6

2 6 Sample Output

6 HINT

The best slice F is f (1,1) =f (2,1) =2,f (1,2) =f (2,2) =1 The idea of solving problems:

Consider the minimum cut.
First of all, it must be (i,j,k) to (i,j,k+1) the edge of the F (i,j,k).
How to satisfy a condition that is less than D. Which is to cut (i,j,k) This side, a side of (I ', J ', k-d~k+d) must be cut off, so (i,j,k) to (i ', J ', k-d), (i ', J ', k+d+1) to the side of the INF (i ', J ', k+1).

#include <bits/stdc++.h> using namespace std;
    int Getint () {int I=0,f=1;char C; For (C=getchar ();(c!= '-') && (c< ' 0 ' | |
    C> ' 9 '); C=getchar ());
    if (c== '-') F=-1,c=getchar ();
    for (; c>= ' 0 ' &&c<= ' 9 '; C=getchar ()) i= (i<<3) + (i<<1) +c-' 0 '; 
return i*f;
const int n=100005,m=1000005,inf=0x3f3f3f3f;
const int fx[4]={-1,0,1,0};
const int fy[4]={0,-1,0,1};
int p,q,r,d,s,t,n,p[45][45][45];
int tot=1,dis[n],cur[n],first[n],nxt[m],to[m],cap[m];

queue<int>q;
    void Add (int x,int y,int z) {nxt[++tot]=first[x],first[x]=tot,to[tot]=y,cap[tot]=z;
nxt[++tot]=first[y],first[y]=tot,to[tot]=x,cap[tot]=0;
    BOOL BFs () {while (!q.empty ()) Q.pop ();
    for (int i=s;i<=t;i++) dis[i]=-1,cur[i]=first[i];
    Q.push (S), dis[s]=0;
        while (!q.empty ()) {int U=q.front (); Q.pop ();
            for (int e=first[u];e;e=nxt[e]) {int v=to[e]; if (Dis[v]==-1&&cap[e]) {Dis[v]=dis[U]+1,q.push (v);
            if (v==t) return true;
}} return false;
    int dinic (int u,int flow) {if (u==t) return flow;
    int res=0;
        for (int &e=cur[u];e;e=nxt[e]) {int v=to[e];
            if (Dis[v]==dis[u]+1&&cap[e]) {int det=dinic (v,min (flow-res,cap[e)));
            Cap[e]-=det,cap[e^1]+=det;
        Res+=det;if (Res==flow) break;
    } if (Res<flow) dis[u]=-1;
return res;
    int main () {//freopen ("lx.in", "R", stdin);
    P=getint (), Q=getint (), R=getint (), D=getint ();
    for (int i=1;i<=p;i++) for (int j=1;j<=q;j++) for (int k=1;k<=r;k++) p[i][j][k]=++n;
    S=0,t=++n;
                for (int k=1;k<=r;k++) for (int i=1;i<=p;i++) for (int j=1;j<=q;j++) {
                int Z=getint ();
                if (k==1) Add (S,p[i][j][k],inf);
                K==r?add (p[i][j][k],t,z): Add (p[i][j][k],p[i][j][k+1],z); for (iNT l=0;l<4;l++) {int di=i+fx[l],dj=j+fy[l]; if (di<1| | di>p| | dj<1| |
                    DJ&GT;Q) continue;
                K>d?add (P[i][j][k],p[di][dj][k-d],inf): Add (P[i][j][k],p[di][dj][1],inf);
    } int ans=0;
    while (BFS ()) ans+=dinic (S,inf);
    cout<<ans;
return 0; }

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.