AcWing 252. Tree (dotted rule) punch

Topic: https://www.acwing.com/problem/content/254/

Meaning of the questions: find a tree, path <= k of how many

Thinking: dotted rule, we solve the two pointers operator functions, firstly calculated the path of each row of a sequence, we ensure monotonic, then l from front to back, from the R & lt forward, if l + r <= m then (l, r-1) ( l, r-2) ... are possible, it can be directly coupled with the total number, and if not r--, satisfies l ++, this example can write their own understandable

 

#include <bits / STDC ++ H.>
 #define MAXN 100005
 #define MOD 1,000,000,007
 the using  namespace STD; 
typedef Long  Long LL; 
LL DA; 
Vector <pair <LL, LL>> MP [MAXN]; // save enough FIG 
BOOL VIS [MAXN]; // tag have used gravity 
LL MAXSIZE [MAXN], DIS [MAXN], D [MAXN]; // MAXSIZE largest subtree of the current node 
LL SIZ [MAXN], E [MAXN]; // dis the distance to the center of gravity d distance appeared 
LL n-, m, rt, SUM, QE;   // distance rt representative of the number of the current node subtree e siz gravity current occurring 
void Find (X LL, LL F) { / / find the center of gravity of 
    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 VIS || F = [q.first]) Continue ; // VIS array tags 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 by subtracting the current number of the current subtree = is the root node of the tree the number of father idea 
    IF (MAXSIZE [X] < MAXSIZE [RT]) { 
        RT =x;
    } 
}
void get_dis(ll x,ll f,ll len){
    if(len<=1e7){
        e[++qe]=len;
    } 
    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;
        get_dis(q.first,x,len+q.second);
    }    
}
ll solve(ll x,ll len){
    ll ee=0;
    qe=0;
    dis[x]=len;
    get_dis(x,0,len);
    sort(e+1,e+qe+1);
    ll l=1,r=qe;
    while(l<r){
        if(e[l]+e[r]<=m){
            ee+=r-l;
            l++;
        }
        else{
            r--;
        }
    }
    return ee;
}
void divide(ll x){
    da+=solve(x,0);
    vis[x]=1;
    for(int i=0;i<mp[x].size();i++){
        pair<ll,ll> q=mp[x][i];
        if(vis[q.first]) continue;
        da-=solve(q.first,q.second);
        sum=siz[q.first];
        rt=0;
        maxsize[rt]=mod;
        find(q.first,x);
        divide(rt);
    }
}
void init(){
    da=0;
    for(int i=0;i<=n;i++) mp[i].clear();
    memset(maxsize,0,sizeof(maxsize));
    memset(vis,0,sizeof(vis)); 
} 
int main(){
    while(cin>>n>>m)
    {
        if(n==0&&m==0) break;
        ll a,b,c;
        init();
        for(int i=0;i<n-1;i++){
            cinA >> B >> >> C; 
            A ++ ; 
            B ++ ; 
            MP [A] .push_back (the make_pair (B, C)); 
            MP [B] .push_back (the make_pair (A, C)); 
        } 
        SUM n-=; // current number of nodes 
        RT = 0 ; 
        MAXSIZE [ 0 ] = MOD; // set the initial value 
        Find ( . 1 , 0 ); 
        Divide (RT); 
        the printf ( " % LLD \ n- " , DA); 
    } 
}