BZOJ2599 [IOI2011] [Race] 点分治

题意：给一棵树,每条边有权.求一条简单路径,权值和等于K,且边的数量最小

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solution：点分治

#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;}