BZOJ 2152 (dotted rule)

Topic: https://vjudge.net/contest/307753#problem/G

Meaning of the questions: a tree allows you to find the total number of paths and 3% of the number of paths

Ideas: dotted flood problem

#include <cstdio> 
#include <CString> 
#include <the cmath> 
#include <algorithm> 
#include <the iostream>  
#include <Vector> 
#include <Queue>
 #define MAXN 100005
 #define MOD 0x3f3f3f3f
 the using  namespace STD; 
typedef Long  Long ll; 
ll DA; 
Vector <pair <ll, ll>> MP [MAXN], XX [MAXN]; // save enough FIG 
ll E [MAXN];
 BOOL VIS [MAXN]; // tag have used gravity 
ll MAXSIZE [MAXN], DIS [MAXN], D [MAXN], in Flag [MAXN]; // MAXSIZE largest subtree of the current node
SIZ LL [MAXN], xD [MAXN]; // DIS distance to the center of gravity d distance appeared 
LL n-, m, K, RT, SUM, QE, QE2, ANS1, ANS2;   // SIZ current node subtree e rt appears from the number of current focus 
void find (X LL, LL F) { // find the center of gravity 
    SIZ [X] = . 1 ; 
    MAXSIZE [X] = 0 ;
     for ( int I = 0 ; I <MP [ X] .size (); I ++ ) { 
        pair <LL, LL> Q = MP [X] [I];
         IF (q.first == F || VIS [q.first]) Continue ; // VIS array tags We have used gravity 
        Find (q.first, X); 
        SIZ [X] + = SIZ [q.first];
        MAXSIZE [X] = max (MAXSIZE [X], SIZ [q.first]); 
    } 
    MAXSIZE [X] = max (MAXSIZE [X], SUM-SIZ [X]); // total number of nodes minus the current sub the number of tree = the current node is the root of the father idea number tree 
    IF (MAXSIZE [X] < MAXSIZE [RT]) { 
        RT = X; 
    } 
} 
void Query (LL Z, LL SM) {
     IF (Z> ANS1) { 
        ANS1 = Z; 
        ANS2 = SM; 
    } 
    the else  IF (Z == ANS1) { 
        ANS2 + = SM; 
    } 
} 
void get_dis (X LL, F LL, LL len) { 
    
    IF (len% . 3==0) ans1++;
    ans2++;
    ans1+=flag[(3-len%3+3)%3];
    ans2+=flag[0]+flag[1]+flag[2];
    e[len%3]++;
    for(int i=0;i<mp[x].size();i++){
        pair<ll,ll> q=mp[x][i];
        if(q.first==f||vis[q.first]) continue;
        //dis[q.first]=(dis[x]+len)%3;
        get_dis(q.first,x,(len+q.second)%3);
    }    
}
void divide(ll x){
    vis[x]=1;
    //printf("rt=%lld ans1=%lld ans2=%lld\n",x,ans1,ans2);
    for(int i=0;i<mp[x].size();i++){
        pair<ll,ll> q=mp[x][i];
        if(vis[q.first]) continue;
        //dis[x]=q.second;
        get_dis(q.first,x,q.second%3);
        for(int j=0;j<3;j++){
            flag[j]+=e[j];
            e[j]=0;
        }
    }
    for(int i=0;i<3;i++){
        flag[i]=0;
    }
    for(int i=0;i<mp[x].size();i++){
        pair<ll,ll> q=mp[x][i];
        if(vis[q.first]) continue;
        //if(da>0) break;
        sum=siz[q.first];
        rt=0;
        maxsize[rt]=mod;
        find(q.first,x);
        divide(rt);
    }
//    vis[x]=0;
}
void init(){
    ans1=0;ans2=0;
    for(int i=0;i<=n;i++) mp[i].clear();
    for(int i=0;i<=n;i++) vis[i]=0;
    for(int i=0;i<3;i++) flag[i]=0;
} 
int main(){
    ll t;
    while(scanf("%lld",&n)!=EOF){
        ll a,b,c;
        init();
        for(int i=1;i<=n-1;i++){
            scanf("%lld%lld%lld",&a,&b,&c);
            mp[a].push_back(make_pair(b,c));
            mp[b].push_back(make_pair(a,c)); 
        }    
        sum=n;//当前节点数 
        rt=0;
        maxsize[0]=mod;//置初值 
        find(1,0);
        divide(rt);
        ans1*=2;
        ans2*=2;
        ans1+=n;
        ans2+=n;
        ll w=__gcd(ans1,ans2);
        //printf("%lld/%lld\n",ans1,ans2);
        printf("%lld/%lld\n",ans1/w,ans2/w);
    
    }
}