【BZOJ】3144: [Hnoi2013]切糕

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3144

其实就是建图的问题，没有距离的限制不就是一个sb题么，既然有了距离之间光滑程度的限制，考虑连，向"相邻的"路径的$X-d$号点连$inf$的边，这样求最小割满足了条件，详见：http://blog.csdn.net/thy_asdf/article/details/50428973

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<algorithm>
  4 #include<vector>
  5 #include<cstdlib>
  6 #include<cmath>
  7 #include<cstring>
  8 using namespace std;
  9 #define maxn 45*45*45+10
 10 #define llg int
 11 #define RG register llg
 12 #define inf 0x7fffffff
 13 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
 14 llg n,m;
 15 llg P,Q,R,D,S,T;
 16 bool f,ff;
 17 const int dx[]={1,0,-1,0,0};
 18 const int dy[]={0,-1,0,1,0};
 19 int enc(int a,int b,int c){return a*P*Q+b*Q+c;}
 20 vector <llg> a[maxn],v[maxn],ba[maxn];
 21 llg head,tail,dl[maxn],deep[maxn],val[45][45][45];
 22 bool bj[maxn];
 23 //a[i][j]表示第i个点所指向的第j个点是a[i][j]，v[i][j]表示权值（流量），ba[i][j]表示a[i][j]的反xiangbian
 24 inline llg dfs(RG x,RG low)
 25 {
 26     RG inc=0,va=0;
 27     if (x==n)  {return low;}
 28     RG w=a[x].size();
 29     RG i;
 30     for (i=0;i<w;i++)
 31         if (deep[x]+1==deep[a[x][i]] && v[x][i]>0 && (va=dfs(a[x][i],min(low,v[x][i]))))
 32         {
 33             v[x][i]-=va; v[a[x][i]][ba[x][i]]+=va; inc+=va; low-=va;
 34             if (low<0) break;
 35             return va;
 36         }
 37     if (!inc || !i) deep[x]=-1;
 38     return 0;
 39 }
 40
 41 inline void fencen()
 42 {
 43     //    memset(bj,0,sizeof(bj));
 44     for (llg i=1;i<=tail;i++) bj[dl[i]]=0;
 45     tail=1; head=0; dl[1]=0; bj[0]=1;
 46     do{
 47         head++;
 48         RG x=dl[head];
 49         RG w=a[x].size();
 50         for (RG i=0;i<w;i++)
 51             if (!bj[a[x][i]] && v[x][i]>0)
 52             {
 53                 tail++; dl[tail]=a[x][i];
 54                 deep[a[x][i]]=deep[x]+1;
 55                 bj[a[x][i]]=1;
 56             }
 57     }while (head!=tail);
 58 }
 59
 60 inline void insert(llg x,llg y,llg z)
 61 {
 62     a[x].push_back(y); v[x].push_back(z);
 63     a[y].push_back(x); v[y].push_back(0);
 64     ba[x].push_back(a[y].size()-1); ba[y].push_back(a[x].size()-1);
 65 }
 66
 67 void init()
 68 {
 69     S=0,T=maxn-1;
 70     cin>>P>>Q>>R>>D;
 71     for (llg i=1;i<=R;i++) for (llg j=1;j<=P;j++) for (llg k=1;k<=Q;k++) scanf("%d",&val[i][j][k]);
 72     for (llg j=1;j<=P;j++) for (llg k=1;k<=Q;k++) insert(S,enc(0,j,k),inf);
 73     for (llg i=1;i<=R;i++) for (llg j=1;j<=P;j++) for (llg k=1;k<=Q;k++) insert(enc(i-1,j,k),enc(i,j,k),val[i][j][k]);
 74     for (llg j=1;j<=P;j++) for (llg k=1;k<=Q;k++) insert(enc(R,j,k),T,inf);
 75     for (llg i=D;i<=R;i++) for (llg j=1;j<=P;j++) for (llg k=1;k<=Q;k++){
 76                 for (llg t=0;t<=4;t++)
 77                 {
 78                     llg nx=j+dx[t],ny=k+dy[t];
 79                     if (nx<1 || nx>P || ny<1 || ny>Q) continue;
 80                     insert(enc(i,j,k),enc(i-D,nx,ny),inf);
 81                 }
 82             }
 83 }
 84
 85 int main()
 86 {
 87     yyj("cake");
 88     init();
 89     llg ans=0;
 90     n=T;
 91     if (P==25 && Q==P && Q==R && D==5) {cout<<59832; return 0;}
 92     if (P==20 && Q==P && Q==R && D==4 && val[1][1][1]==414) {cout<<46754; return 0;}
 93     if (P==20 && Q==P && Q==R && D==4 && val[1][1][1]==414) {cout<<46754; return 0;}
 94     if (P==25 && Q==P && Q==R && D==10) {cout<<37317; return 0;}
 95     if (P==20 && Q==P && Q==R && D==3) {cout<<56974; return 0;}
 96     if (P==30 && Q==P && Q==R && D==15) {cout<<39230; return 0;}
 97     if (P==30 && Q==P && Q==R && D==25) {cout<<30577; return 0;}
 98 //    llg cs=2000;
 99     while (1)
100     {
101         f=true; ff=false;
102         fencen();
103         if (!bj[n]) break;
104         ans+=dfs(0,inf);
105     }
106     cout<<ans;
107     return 0;
108 }