0-1-Tree CodeForces

大意: 给定树, 边权为黑或白, 求所有有向路径条数, 满足每走过一条黑边后不会走白边.

这题比赛的时候想了个假算法, 还没发现..... 

显然所求的路径要么全黑, 要么全白, 要么先全白后全黑, 所以可以用并查集将相邻同色边合并即可.

#include <iostream>
#include <random>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head


const int N = 1e6+10;
int n, sz1[N], sz2[N], fa1[N], fa2[N];
int f1(int x) {return fa1[x]?fa1[x]=f1(fa1[x]):x;}
int f2(int x) {return fa2[x]?fa2[x]=f2(fa2[x]):x;}
void add1(int x, int y) {if ((x=f1(x))!=(y=f1(y))) fa1[x]=y,sz1[y]+=sz1[x];}
void add2(int x, int y) {if ((x=f2(x))!=(y=f2(y))) fa2[x]=y,sz2[y]+=sz2[x];}
int main() {
	scanf("%d", &n);
	REP(i,1,n) sz1[i]=sz2[i]=1;
	REP(i,2,n) {
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		w?add2(u,v):add1(u,v);
	}
	ll ans = 0;
	REP(i,1,n) ans+=(ll)sz1[f1(i)]*sz2[f2(i)]-1;
	printf("%lld
", ans);
}