http://codeforces.com/contest/1118/problem/F1
题意:给一棵树,有些节点是红色或蓝色,或无色,切掉一条边后分成两棵树且红蓝在不同的树;
有多少条边可以切;
思路:一个节点的一个子树点有么包含红的全部且不包含蓝的,或包含蓝的全部而不包含红的,那么这条边可以切;dfs是计算子树红蓝个数是否满足条件;
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;
#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=3e5+9;
const int mod=1e9+7;
template <class T>
inline void sc(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
vector<int>g[maxn];
int a[maxn];
int ans=0,red=0,blue=0;
pair<int,int> dfs(int u,int fa)
{
int r=0;
int b=0;
if(a[u]==1) r++;
if(a[u]==2) b++;
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(v==fa) continue;
pair<int,int> tmp=dfs(v,u);
if(tmp.first==red&&tmp.second==0) ans++;
if(tmp.first==0&&tmp.second==blue) ans++;
r+=tmp.first;
b+=tmp.second;
}
return make_pair(r,b);
}
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]==1) red++;
if(a[i]==2) blue++;
}
for(int i=0;i<n-1;i++)
{
int u,v;
cin>>u>>v;
u--;
v--;
g[u].push_back(v);
g[v].push_back(u);
}
ans=0;
dfs(0,-1);
cout<<ans<<endl;
return 0;
}