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_queuetypedef 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; //
const ll nmos = 0x80000000LL; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //
const int mod = ;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+;
int root = ,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 = ;
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] = ,f[u] = ;
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 = ] = D;
getDis(x,,D);
sort(dis+,dis++cnt);
int le = ,ri =cnt,ans = ;
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,);
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,-,sizeof(h));
memset(used,false,sizeof(used));
tot = ;
for(int i=; 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(,-);
printf("%d\n",Divide(root));
}
return ;
}
POJ-1741
自己今天又写了一遍。
//点分治
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000") //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>using namespace std;
#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_queuetypedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
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)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;const ll mos = 0x7FFFFFFF; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0);
const double PHI=0.61803399; //黄金分割点
const double tPHI=0.38196601;template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
}/*-----------------------showtime----------------------*/
/*
using namespace std;
#define pb push_back
#define debug(x) cerr<<#x<<" = " << x<<endl;
#define fi first
#define se second
typedef pair<int,int> pii;
const int inf = 0x3f3f3f3f;*/
int n,k,ans;
const int maxn = ;
vector<pii>mp[maxn];
int dp[maxn],cen[maxn];
void getsize(int u,int fa){
dp[u] = ;
for(int i=; i<mp[u].size(); i++){
int v = mp[u][i].fi;
if(fa == v || cen[v]) continue;
getsize(v, u);
dp[u] += dp[v];
}
} pii getbig(int u,int fa,int t){
pii res = pii(inf, u);
int mx = ;
for(int i=; i<mp[u].size(); i++){
int v = mp[u][i].fi;
if(v == fa || cen[v])continue;
res = min(res, getbig(v, u, t));
mx = max(mx, dp[v]);
} res = min(res, pii(max(mx, t - dp[u]), u));
return res;
} void dfs(int u,int fa,int c, vector<int> & b){
b.pb(c);
for(int i=; i<mp[u].size(); i++){
int v = mp[u][i].fi,d = c + mp[u][i].se;
if(v == fa || cen[v])continue;
dfs(v,u,d,b);
}
}
int cal(vector<int>&b){ sort(b.begin(), b.end()); int res = ,r = b.size();
for(int i=; i<b.size(); i++){
while(r && b[i] + b[r-] > k) r--;
if(r > i) res += r - ;
else res += r;
}
return res/;
} void solve(int u){
getsize(u, -);
int s = getbig(u, -,dp[u]).se;
cen[s] = ; for(int i=; i<mp[s].size(); i++){
int v = mp[s][i].fi;
if(cen[v])continue;
solve(v);
} vector<int>a;
a.pb();
for(int i=; i<mp[s].size(); i++)
{
int v = mp[s][i].fi;
if(cen[v])continue; vector<int>b;
dfs(v, s, mp[s][i].se, b);
ans -= cal(b);
a.insert(a.end(),b.begin(),b.end());
} ans += cal(a);
cen[s] = ;
}
int main(){
while(~scanf("%d%d", &n, &k) && n + k){
for(int i=; i<=n; i++) mp[i].clear();
for(int i=; i<n; i++){
int u,v,w;
scanf("%d%d%d", &u, &v, &w);
mp[u].pb(pii(v,w));
mp[v].pb(pii(u,w));
}
ans = ;
solve();
printf("%d\n", ans);
}
return ;
}
new
