题意:
对于带权的一棵树,求树中距离不超过k的点的对数。
思路:
点分治的裸题。 将这棵树分成很多小的树,分治求解。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <cmath> #include <queue> #include <list> #include <map> #include <set> using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ‘\n‘ #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const int mod = 998244353; const double PI=acos(-1.0); // #define _DEBUG; //*// #ifdef _DEBUG freopen("input", "r", stdin); // freopen("output.txt", "w", stdout); #endif /*-----------------------showtime----------------------*/ const int maxn = 1e5+9; int root = 0,S,mx; int n,k; int sz[maxn],f[maxn],dis[maxn],cnt; bool used[maxn]; struct node { int to,w,nx; }e[maxn]; int h[maxn],tot = 0; void add(int u,int v,int w){ e[tot].to = v; e[tot].w = w; e[tot].nx = h[u]; h[u] = tot++; } void getRoot(int u, int fa){ sz[u] = 1,f[u] = 1; for(int i = h[u] ; ~i; i= e[i].nx){ int v = e[i].to; if(used[v] || fa == v)continue; getRoot(v,u); sz[u] += sz[v]; f[u] = max(f[u] , sz[v]); } f[u] = max(f[u],S - sz[u]); if(f[u] < mx){root = u;mx = f[u];} } void getDis(int u,int fa,int D){ for(int i=h[u] ; ~i; i=e[i].nx){ int v = e[i].to; if(used[v]||v == fa)continue; dis[++cnt] = D + e[i].w; getDis(v,u,dis[cnt]); } } int getAns(int x,int D){ dis[cnt = 1] = D; getDis(x,0,D); sort(dis+1,dis+1+cnt); int le = 1,ri =cnt,ans = 0; while(le <= ri){ if(dis[le] + dis[ri] <= k)ans += ri - le,le++; else ri--; } return ans; } int Divide(int x){ used[x] = true; ll ans = getAns(x,0); for(int i=h[x]; ~i; i= e[i].nx){ int v = e[i].to; if(used[v])continue; ans -= getAns(v,e[i].w); mx = inf,S = sz[v]; getRoot(v,x);ans += Divide(root); } return ans; } int main(){ while(~scanf("%d%d", &n, &k) && n+k) { memset(h,-1,sizeof(h)); memset(used,false,sizeof(used)); tot = 0; for(int i=1; i<n; i++){ int u,v,c; scanf("%d%d%d", &u, &v,&c); add(u,v,c); add(v,u,c); } S = n;mx = inf; getRoot(1,-1); printf("%d\n",Divide(root)); } return 0; }
原文:https://www.cnblogs.com/ckxkexing/p/9508386.html