A Pair of Graphs（图的构造） HDU 2464

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本文介绍了一种通过深度优先搜索解决图构造问题的方法，旨在找到使两个图等价所需的最小操作成本。通过调整搜索策略，将算法的时间复杂度显著降低。

A Pair of Graphs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 357 Accepted Submission(s): 184

Problem Description

We say that two graphs are equivalent if and only if a one-to-one correspondence between their nodes can be established and under such a correspondence their edges are exactly the same. Given A and B, two undirected simple graphs with the same number of vertexes, you are to find a series of operations with the minimum cost that will make the two graphs equivalent. An operation may be one of the following two types:
a) Add an arbitrary edge (x, y), provided that (x, y) does not exist before such an operation. Such an operation costs I _A and I _B on two graphs, respectively.
b) Delete an existing edge (x, y), which costs D _A and D _B on two graphs,respectively.

Input

There are multiple test cases in the input file.
Each test case starts with three integers, N, M _A and M _B, ( 1 ≤ N ≤ 8, 0 ≤ M _B ,M _A ≤ N*(N-1)/2 ), the total number of vertexes, the number of edges in graph A,and the number of edges in graph B, respectively. Four integers I _A, I _B, D _A, and D _B come next, (0 ≤ I _A, I _B, D _A, D _B ≤ 32767 ), representing the costs as stated in the problem description. The next M _A + M _B lines describe the edges in graph A followed by those in graph B. Each line consists of exactly two integers, X and Y ( X ≠ Y , 0 ≤ X,Y < N).
Two successive test cases are separated by a blank line. A case with N = 0, M _A = 0,and M _B = 0 indicates the end of the input file, and should not be processed by your program.

Output

Please print the minimum cost in the format as illustrated below.

Sample Input

Sample Output

  
   Case #1: 0
Case #2: 1

Source

2008 Asia Regional Hangzhou

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算法分析

这是图的构造问题。首先先构造好图，点与点是否存在通路，两个数组A,B分别拿来保存是否存在通路，接着用深度搜索出所有可能，对，就是穷搜所有可能性，再查看是否满足条件，由于N<=8,最大不过是8!*N*N，不是特别大了，最大经过一些剪枝，讲时间降到了0.29秒。

#include<stdio.h>
#include<string.h>
bool a[9][9],b[9][9],used[9];
int n,ma,mb,ia,ib,da,db,ii,dd,num,list[9];
int min(int a,int b)
{
if(a>b) return b;
return a;
}
void dfs(int k) //由于N小于等于8，所以可以使用穷搜，为N!，8！= 40320，很小
{
int i,j,num2=0;
if(k==n)
{
  for(i=0;i<n;i++)
   for(j=0;j<n;j++)
   {
   if(a[list[i]][list[j]]==b[i][j]) continue;
   if(a[list[i]][list[j]])
       num2+=dd;//存在A中边不存在B中边说明要么A减少要么B增加
   else
       num2+=ii;//存在B中边不存在A中边说明要么B减少要么A增加
   }
  if(num>num2) num=num2;
  return;
}
else
for(i=0;i<n;i++)
if(!used[i])
{
  list[k]=i;
  used[i]=true;
  dfs(k+1);
  used[i]=false;
}
}
int main()
{
int i,j,x,y,count=0;
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
while(scanf("%d%d%d",&n,&ma,&mb)!=EOF)
{
  scanf("%d%d%d%d",&ia,&ib,&da,&db);
  ii=min(ia,db);//增加A和减少B是等价的，因为我们是以A为研究对象，逐步构成成为B
  dd=min(ib,da);//增加B和减少A是等价的
  if(n==0&&ma==0&&mb==0) break;
  memset(a,false,sizeof(a));
  memset(b,false,sizeof(b));
  memset(used,false,sizeof(used));
  for(i=1;i<=ma;i++)
  {
   scanf("%d%d",&x,&y);
   a[x][y]=a[y][x]=true;
  }
  for(i=1;i<=mb;i++)
  {
   scanf("%d%d",&x,&y);
   b[x][y]=b[y][x]=true;
  }
  num=1<<25;
  dfs(0);
  printf("Case #%d: %d/n",++count,num/2);
}
return 0;
}

改进版，时间从0.50到了0.29，快了很多

#include<stdio.h>
bool a[9][9],b[9][9],used[9];
int n,ma,mb,ia,ib,da,db,ii,dd,num,list[9];
int min(int a,int b)
{
if(a>b) return b;
return a;
}
void dfs(int k) //由于N小于等于8，所以可以使用穷搜，为N!，8！= 40320，很小
{
int i,j,num2=0;
if(k==n)
{
  for(i=0;i<n;i++)
   for(j=i+1;j<n;j++)//这个的改变速度变化很大
   {
   if(a[list[i]][list[j]]==b[i][j]) continue;
   if(a[list[i]][list[j]])
       num2+=dd;//存在A中边不存在B中边说明要么A减少要么B增加
   else
       num2+=ii;//存在B中边不存在A中边说明要么B减少要么A增加
   }
  if(num>num2) num=num2;
  return;
}
else
for(i=0;i<n;i++)
if(!used[i])
{
  list[k]=i;
  used[i]=true;
  dfs(k+1);
  used[i]=false;
}
}
int main()
{
int i,j,x,y,count=0;
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
while(scanf("%d%d%d",&n,&ma,&mb)!=EOF)
{
  scanf("%d%d%d%d",&ia,&ib,&da,&db);
  ii=min(ia,db);//增加A和减少B是等价的，因为我们是以A为研究对象，逐步构成成为B
  dd=min(ib,da);//增加B和减少A是等价的
  if(n==0&&ma==0&&mb==0) break;
  for(i=0;i<n;i++)
  {
    for(j=0;j<n;j++)
    {
    a[i][j]=false;
    b[i][j]=false;
    }
    used[i]=false;
  }
  for(i=1;i<=ma;i++)
  {
   scanf("%d%d",&x,&y);
   a[x][y]=a[y][x]=true;
  }
  for(i=1;i<=mb;i++)
  {
   scanf("%d%d",&x,&y);
   b[x][y]=b[y][x]=true;
  }
  num=1<<25;
  dfs(0);
  printf("Case #%d: %d/n",++count,num);
}
return 0;
}