标签:ORC getc ios add bitset highlight mes div return
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 ‘\n‘ #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\n", ans); }
0-1-Tree CodeForces - 1156D (并查集)
标签:ORC getc ios add bitset highlight mes div return
原文地址:https://www.cnblogs.com/uid001/p/10802874.html