如果原图不连通,直接输出0.

如果原图连通,删除X条边之后要保证新图连通,再看数据是n+1条边-->因此,最多只能删去两条边。

因为n=100,可以枚举进行验证,枚举删去每一条边是否连通,枚举删去每两条边是否连通,验证是否连通可以用并查集,可以BFS。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<queue>
#include<algorithm>
#include<iostream>
using namespace std; const int maxn=+;
int n;
int fa[maxn];
int u[maxn],v[maxn];
long long c[maxn][maxn];
int T;
int ans; int Find(int x)
{
if(x!=fa[x]) fa[x]=Find(fa[x]);
return fa[x];
} int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d",&n); ans=;
for(int i=;i<=n+;i++)
scanf("%d%d",&u[i],&v[i]); for(int i=;i<=n;i++) fa[i]=i;
int un=n;
for(int i=;i<=n+;i++)
{
int fu=Find(u[i]);
int fv=Find(v[i]);
if(fu!=fv)
{
un--;
fa[fu]=fv;
}
} if(un!=) ans=;
else
{
for(int i=;i<=n+;i++)
{
for(int k=;k<=n;k++) fa[k]=k;
un=n;
for(int k=;k<=n+;k++)
{
if(k==i) continue;
int fu=Find(u[k]);
int fv=Find(v[k]);
if(fu!=fv)
{
un--;
fa[fu]=fv;
}
}
if(un==) ans=ans+;
} for(int i=;i<=n+;i++)
{
for(int j=i+;j<=n+;j++)
{
for(int i=;i<=n;i++) fa[i]=i;
un=n;
for(int k=;k<=n+;k++)
{
if(k==i||k==j) continue;
int fu=Find(u[k]);
int fv=Find(v[k]);
if(fu!=fv)
{
un--;
fa[fu]=fv;
}
}
if(un==) ans=ans+;
}
}
}
printf("%d\n",ans);
}
return ;
}

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