POJ1741--Tree （树的点分治） 求树上距离小于等于k的点对数

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12276		Accepted: 3886

Description

Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input

The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
Output

For each test case output the answer on a single line.
Sample Input

5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

Sample Output

8

题意：求树上距离小于等于k的点对数。。

我是用 点的分治做的，，据说还可以用启发式合并做，，附上链接http://blog.csdn.net/asdfgh0308/article/details/39845489。。挖个坑。

定义树的重心 s 为 删除s点后的 最大子树的点数 小于n/2。 那么对于任意满足条件的点对 有两种情况，，

其路径 1要么经过s 2要么不经过s。。

对于1 我们只需要 求出 以s为根的子树的点到s的距离即可。。

对于2 可以递归处理 分解为 多个1。。然后就可以求出来了。。

复杂度为nlogn*logn

#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e4+10;
struct Edge
{
int to, len;
Edge (int _x, int _len)
{
to = _x;
len = _len;
}
};
vector<Edge>G[maxn << 1];
int siz[maxn];
bool vis[maxn];
void Get_size(int root, int father)
{
siz[root] = 1;
for (int i = 0; i < G[root].size(); i++)
{
int v = G[root][i].to;
if (v == father || vis[v] == true)
continue;
Get_size(v, root);
siz[root] += siz[v];
}
}
typedef pair<int,int>pii;
pii FindGravity(int root, int father, int t)
{
pii res = make_pair(inf, -1);
int m = 0, sum = 1;
for (int i = 0; i < G[root].size(); i++)
{
int v = G[root][i].to;
if (v == father || vis[v] == true)
continue;
res = min(res, FindGravity(v, root, t));
m = max(m, siz[v]);
sum += siz[v];
}
m = max(m, t-sum);
res = min(res, make_pair(m, root));
return res;
}
void Get_len(int root, int father, int d, vector<int>&len)
{
len.push_back(d);
for (int i = 0; i < G[root].size(); i++)
{
int v = G[root][i].to;
if (v == father || vis[v] == true)
continue;
Get_len(v, root, d+G[root][i].len, len);
}
}
int K;
int cnt_pair(vector<int>&ds)
{
int res = 0;
sort (ds.begin(), ds.end());
int j = ds.size() - 1;
for (int i = 0; i < ds.size(); i++)
{
while (j > i && ds[i] + ds[j] > K)
{
j--;
}
res += (j > i ? j - i : 0);
}
return res;
}
int solve(int root)
{
int ans = 0;
Get_size(root, -1);
int s = FindGravity(root, -1, siz[root]).second;
if (s == -1)
return 0;
vis[s] = true;
for (int i = 0; i < G[s].size(); i++)
{
int v = G[s][i].to;
if (vis[v] == true)
continue;
ans += solve(v);
}
vector<int>ds;
ds.push_back(0);
for (int i = 0; i < G[s].size(); i++)
{
int v = G[s][i].to;
if (vis[v] == true)
continue;
vector<int>rds;
Get_len(v, s, G[s][i].len, rds);
ans -= cnt_pair(rds);
ds.insert(ds.end(), rds.begin(), rds.end());
}
ans += cnt_pair(ds);
vis[s] = false;
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int n;
while ( scanf ("%d%d", &n, &K), n && K)
{
int u, v, c;
for (int i = 0; i <= n; i++)
G[i].clear();
for (int i = 0; i < n-1; i++)
{
scanf ("%d%d%d", &u, &v, &c);
G[u].push_back(Edge(v,c));
G[v].push_back(Edge(u,c));
}
printf("%d\n", solve(n/2+1));
}
return 0;
}

②第二种姿势，，200ms左右

#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e4+10;
struct Edge
{
int to, len, next;
}e[maxn << 1];
int head[maxn], tot, N, K;
void add_edge(int u, int v, int c)
{
e[tot].to = v;
e[tot].len = c;
e[tot].next = head[u];
head[u] = tot++;
}
int siz[maxn],Mtree[maxn];  // Mtree为最大子树的大小 siz为子树的大小
bool vis[maxn];
int center;
void FindGravity(int u, int father, int cnt)  // 查找重心 center
{
siz[u] = 1;
Mtree[u] = 0;
for (int i = head[u]; ~i; i = e[i].next)
{
int v = e[i].to;
if (v == father || vis[v] == true)
continue;
FindGravity(v, u, cnt);
siz[u] += siz[v];
Mtree[u] = max(Mtree[u], siz[v]);
}
Mtree[u] = max(cnt - siz[u], Mtree[u]);
if (Mtree[center] > Mtree[u])
center = u;
}
int S[maxn], dep[maxn], top;
void Get_len(int u, int father, int d)
{
S[top++] = d;
for (int i = head[u]; ~i; i = e[i].next)
{
int v = e[i].to;
if (vis[v] == true || v == father)
continue;
//  dep[v] = dep[u] + e[i].len;
Get_len(v, u, d+e[i].len);
}
}
int Get_cnt(int u, int d)
{
int res = 0;
top = 0;
//dep[u] = d;
Get_len(u, 0, d);
sort (S, S+top);
int j = top - 1;
for (int i = 0; i < top; i++)
{
while (j > i && S[i] + S[j] > K)
j--;
res += (j > i ? j-i : 0);
}
return res;
}
int ans;
void solve(int r)
{
vis[r] = true;
ans += Get_cnt(r, 0);
for (int i = head[r]; ~i; i = e[i].next)
{
int v = e[i].to;
if (vis[v] == true)
continue;
ans -= Get_cnt(v, e[i].len);
center = 0;
FindGravity(v, r, siz[v]);
solve(center);
}
}
void init()
{
tot = 0;
Mtree[0] = N;
memset(head, -1, sizeof(head));
memset(vis, false, sizeof(vis));
center = 0;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
while (scanf ("%d%d", &N,&K), N && K)
{
init();
int u, v, c;
for (int i = 0; i < N-1; i++)
{
scanf ("%d%d%d", &u, &v, &c);
add_edge(u, v, c);
add_edge(v, u, c);
}
ans = 0;
FindGravity(N/2+1, -1, N);
solve(center);
printf("%d\n", ans);
}
return 0;
}