匈牙利算法来找最多的没有弱化的！！！！但是初始化那块不知道为什么换了个地方就re。。。尴尬

#include<cstring>
#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

int a[15];
int G[15][15];
int g[10][10];

int n;
int linker[15];
bool used[15];

bool dfs(int u)
{
for(int v=1; v<=n; v++)
{
if(g[u][v]&&!used[v])
{
used[v]=true;
if(linker[v]==0||dfs(linker[v]))
{
linker[v]=u;
return true;
}
}
}
return false;
}

int hungary()
{
int res=0;
memset(linker,0,sizeof linker);//难道非得在这里初始化么？
for(int u=1; u<=n; u++)
{
memset(used,false,sizeof used);
if(dfs(u))
res++;
}
return res;
}

int main()
{
int m;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0)
{
printf("0\n");
continue;
}
memset(G,0,sizeof G);
for(int i=1; i<=m; i++)
{
int a,b;
scanf("%d%d",&a,&b);
G[a]=1;
}
int ans=100;
for(int i=1; i<=n; i++)
a[i]=i;
do
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
g[i][j]=0;／／这里也是必须在里面一个一个初始化，不能直接在外面ｍｅｍｓｅｔ
if(!G[i][a[j]]&&!G[i][a[j%n+1]])
g[i][j]=1;
}
ans=min(ans,n-hungary());
}while(next_permutation(a+2,a+n+1));
printf("%d\n",ans);
}
return 0;
}

Necklace

[b][b]Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1111    Accepted Submission(s): 361
[/b]

[align=left]Problem Description[/align]
SJX has 2*N magic gems.
N
of them have Yin energy inside while others have Yang energy. SJX wants to make a necklace with these magic gems for his beloved BHB. To avoid making the necklace too Yin or too Yang, he must place these magic gems Yin after Yang and Yang after Yin, which
means two adjacent gems must have different kind of energy. But he finds that some gems with Yang energy will become somber adjacent with some of the Yin gems and impact the value of the neckless. After trying multiple times, he finds out M rules of the gems.
He wants to have a most valuable neckless which means the somber gems must be as less as possible. So he wonders how many gems with Yang energy will become somber if he make the necklace in the best way.
 

[align=left]Input[/align]
  Multiple test cases.

  For each test case, the first line contains two integers
N(0≤N≤9),M(0≤M≤N∗N),
descripted as above.

  Then M
lines followed, every line contains two integers X,Y,
indicates that magic gem X
with Yang energy will become somber adjacent with the magic gem
4000

Y
with Yin energy.

 

[align=left]Output[/align]
One line per case, an integer indicates that how many gem will become somber at least.
 

[align=left]Sample Input[/align]

2 1
1 1
3 4
1 1
1 2
1 3
2 1

 

[align=left]Sample Output[/align]

1
1