Petya and Spiders DLX+重复覆盖最小支配集(最少多少个点能覆盖所有的点)

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本文深入探讨了游戏开发领域的核心技术，包括游戏引擎、图形渲染、3D空间视频等关键概念及应用实例。

Little Petya loves training spiders. Petya has a board n × m in size. Each cell of the board initially has a spider sitting on it. After one second Petya chooses a certain action for each spider, and all of them humbly perform its commands. There are 5 possible commands: to stay idle or to move from current cell to some of the four side-neighboring cells (that is, one command for each of the four possible directions). Petya gives the commands so that no spider leaves the field. It is allowed for spiders to pass through each other when they crawl towards each other in opposite directions. All spiders crawl simultaneously and several spiders may end up in one cell. Petya wants to know the maximum possible number of spider-free cells after one second.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 40, n·m ≤ 40) — the board sizes.

Output

In the first line print the maximum number of cells without spiders.

Sample test(s)

Input

1 1

Output

Input

2 3

Output

Note

In the first sample the only possible answer is:

In the second sample one of the possible solutions is:

rdl
rul

s denotes command "stay idle", l, r, d, u denote commands "crawl left", "crawl right", "crawl down", "crawl up", correspondingly.

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define eps 1e-8
#define N 185
#define V 36000
int n,m;//n行 m列
int L[V],R[V];
int D[V],U[V];
int C[V];
int S[N],H[N],mark[V],OK[N],tans[V];
int ak,size;//ak 最少多少行可以覆盖所有列(可重复)
void Link(int r,int c)
{
S[c]++;C[size]=c;
U[size]=U[c];D[U[c]]=size;
D[size]=c;U[c]=size;
if(H[r]==-1) H[r]=L[size]=R[size]=size;
else
{
L[size]=L[H[r]];R[L[H[r]]]=size;
R[size]=H[r];L[H[r]]=size;
}
mark[size]=r;
size++;
}
void remove(int c)
{
int i;
for(i=D[c];i!=c;i=D[i])
L[R[i]]=L[i],R[L[i]]=R[i];
}
void resume(int c)
{
int i;
for(i=U[c];i!=c;i=U[i])
L[R[i]]=R[L[i]]=i;
}
int h()
{
int i,j,k,count=0;
bool visit[N];
memset(visit,0,sizeof(visit));
for(i=R[0];i;i=R[i])
{
if(visit[i]) continue;
count++;
visit[i]=1;
for(j=D[i];j!=i;j=D[j])
{
for(k=R[j];k!=j;k=R[k])
visit[C[k]]=1;
}
}
return count;
}
void Dance(int k)
{
int i,j,c,Min,ans;
ans=h();
if(k+ans>=ak) return;
if(!R[0])
{
if(k<ak)
{
ak=k;
for(int i=0;i<k;i++)
{
tans[i]=mark[OK[i]];
}
}
return;
}
for(Min=N,i=R[0];i;i=R[i])
if(S[i]<Min) Min=S[i],c=i;
for(i=D[c];i!=c;i=D[i])
{
OK[k]=i;
remove(i);
for(j=R[i];j!=i;j=R[j])
remove(j);
Dance(k+1);
for(j=L[i];j!=i;j=L[j])
resume(j);
resume(i);
}
return;
}
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
int mat[400][400];
int main()
{
int tn,tm;
while(scanf("%d%d",&tn,&tm)==2)
{
memset(mat,0,sizeof(mat));
for(int i=1;i<=tn;i++)
{
for(int j=1;j<=tm;j++)
{
for(int k=0;k<5;k++)
{
int x=i+xx[k],y=j+yy[k];
if(x>=1&&x<=tn&&y>=1&&y<=tm)
{
mat[(i-1)*tm+j][(x-1)*tm+y]=1;
}
}
}
}
n=tn*tm,m=tn*tm;
//DLX
for(int i=0;i<=m;i++)
{
S[i]=0;
U[i]=D[i]=i;
L[i+1]=i;R[i]=i+1;
}R[m]=0;
memset(H,-1,sizeof(H));
memset(mark,0,sizeof(mark));
memset(tans,0,sizeof(tans));
size=m+1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{//cout<<i<<".."<<j<<endl;
if(mat[i][j]) Link(i,j);
}
}
ak=N;
Dance(0);
printf("%d\n",n-ak);
// for(int i=0;i<ak;i++) cout<<tans[i]<<endl;
}
return 0;
}