BZOJ2599 [IOI2011] [Race] 點分治

題意：給一棵樹,每條邊有權.求一條簡單路徑,權值和等於K,且邊的數量最小

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solution：點分治

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 2e5 + 7;
const int Inf = 2000000000;
struct Edge{
    int v , w;
    Edge ( int v = 0 , int w = 0 ) : v(v) , w(w) {}
};
vector <Edge> edges;
vector <int> G[maxn*2];

void addeage ( int u , int v , int w ){
    edges.push_back ( Edge ( v , w ) );
    int tot = edges.size();
    G[u].push_back ( tot - 1 );
}

struct P{
    int d, p;
}a[maxn];

int root , mxrt , sum , cnt , n , K;
int vis[maxn] , son[maxn];
int dep[maxn], dis[maxn];
int ans[maxn];

void getroot( int u , int fa ){
    son[u] = 1;
    int mx = 0;
    for ( int i = 0 ; i < G[u].size() ; ++ i ){
        Edge e = edges[ G[u][i] ];
        if ( vis[e.v] != 0 || e.v == fa ) continue;
        getroot ( e.v , u );
        son[u] += son[e.v];
        mx = max ( mx , son[e.v] );
    }
    mx = max ( mx , sum - son[u] );
    if ( mx < mxrt ){
        root = u;
        mxrt = mx;
    }
}

void getdis ( int u , int fa ){
    a[++cnt].d = dis[u] , a[cnt].p = dep[u];
    for ( int i = 0 ; i < G[u].size() ; ++ i ){
        Edge e = edges[G[u][i]];
        if ( e.v == fa || vis[e.v] ) continue;
        dep[e.v] = dep[u] + 1;
        dis[e.v] = dis[u] + e.w;
        getdis( e.v , u );  
    }   
}

int cmp ( P x , P y ){
    return x.d < y.d;
}
void cal ( int u , int ndis , int ndep , int delta ){
    dis[u] = ndis , dep[u] = ndep , cnt = 0;
    getdis ( u , -1 );
    sort ( a + 1 , a + 1 + cnt , cmp );
    int l = 1 , r = cnt ;
    while ( l <= r ){
        while ( l<r && a[l].d + a[r].d > K ) -- r;
        int i = r;
        while ( a[l].d + a[i].d == K ){
//          printf ( "find = %d %d %d %d\n" , a[l].d , a[i].d , a[l].p , a[i].p );
            ans[ a[l].p + a[i].p ] += delta;
            -- i;
        }
        ++ l;
    }
}

void work ( int x ){
//  printf ( "root = %d\n" , x );
    vis[x] = 1;
    cal ( x , 0 , 0 , 1 );
    for ( int i = 0 ; i < G[x].size() ; ++ i ){
        Edge e = edges[G[x][i]];
        if ( vis[e.v] ) continue;
        cal ( e.v , e.w , 1 , -1 );
        root = 0;
        mxrt = Inf;
        sum = son[e.v];
        getroot( e.v , -1 );
        work ( root );
    }
}

inline int readin(){
    int x = 0 , f = 1; char ch = getchar();
    while ( !isdigit(ch) ) {
        if ( ch == '-' ) f = -1;
        ch = getchar();
    }
    while ( isdigit(ch) ){
        x = x * 10 + ch - '0' ;
        ch = getchar();
    }
    return x*f;
}
int main(){
//  freopen( "debug.in" , "r" , stdin );
//  scanf ( "%d%d" , &n , &K );
    n = readin(), K = readin();
//  memset( vis , 0 , sizeof(vis) );
    for ( int i = 1 ; i < n ; ++ i ){   
        int x , y , z;
        x = readin(), y = readin(), z = readin();
//      scanf ( "%d%d%d" , &x , &y , &z );
        ++ x , ++ y ;
    //  printf ( "%d %d\n" , x , y );
        addeage ( x , y , z );
        addeage ( y , x , z );
    }
/*  for ( int i = 1 ; i <= n ; ++ i ){
        printf ( "%d : ", i );
        for ( int j = 0 ; j < G[i].size() ; ++ j )
            printf ( " %d" , edges[G[i][j]].v );
        puts ( "" );
    }
*/  mxrt = Inf;
    sum = n;
    root = 0;
    getroot ( 1 , -1 );
//  printf ( "root is %d \n" , root );
    work ( root );
    for ( int i = 1 ; i <= n ; i++ ) 
        if ( ans[i] != 0 ){
            printf ( "%d" , i );
            return 0;
        }
    printf ( "-1" );
    return 0;
}