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

  1 #include <set>

  2 #include <map>

  3 #include <cmath>

  4 #include <ctime>

  5 #include <queue>

  6 #include <stack>

  7 #include <cstdio>

  8 #include <string>

  9 #include <vector>

 10 #include <cstdlib>

 11 #include <cstring>

 12 #include <iostream>

 13 #include <algorithm>

 14 using namespace std;

 15 typedef unsigned long long ull;

 16 typedef long long ll;

 17 const int inf = 0x3f3f3f3f;

 18 const double eps = 1e-8;

 19 const int maxn = 1e4+10;

 20 struct Edge

 21 {

 22     int to, len;

 23     Edge (int _x, int _len)

 24     {

 25         to = _x;

 26         len = _len;

 27     }

 28 };

 29 vector<Edge>G[maxn << 1];

 30 int siz[maxn];

 31 bool vis[maxn];

 32 void Get_size(int root, int father)

 33 {

 34     siz[root] = 1;

 35     for (int i = 0; i < G[root].size(); i++)

 36     {

 37         int v = G[root][i].to;

 38         if (v == father || vis[v] == true)

 39             continue;

 40         Get_size(v, root);

 41         siz[root] += siz[v];

 42     }

 43 }

 44 typedef pair<int,int>pii;

 45 pii FindGravity(int root, int father, int t)

 46 {

 47     pii res = make_pair(inf, -1);

 48     int m = 0, sum = 1;

 49     for (int i = 0; i < G[root].size(); i++)

 50     {

 51         int v = G[root][i].to;

 52         if (v == father || vis[v] == true)

 53             continue;

 54         res = min(res, FindGravity(v, root, t));

 55         m = max(m, siz[v]);

 56         sum += siz[v];

 57     }

 58     m = max(m, t-sum);

 59     res = min(res, make_pair(m, root));

 60     return res;

 61 }

 62 void Get_len(int root, int father, int d, vector<int>&len)

 63 {

 64     len.push_back(d);

 65     for (int i = 0; i < G[root].size(); i++)

 66     {

 67         int v = G[root][i].to;

 68         if (v == father || vis[v] == true)

 69             continue;

 70         Get_len(v, root, d+G[root][i].len, len);

 71     }

 72 }

 73 int K;

 74 int cnt_pair(vector<int>&ds)

 75 {

 76     int res = 0;

 77     sort (ds.begin(), ds.end());

 78     int j = ds.size() - 1;

 79     for (int i = 0; i < ds.size(); i++)

 80     {

 81         while (j > i && ds[i] + ds[j] > K)

 82         {

 83             j--;

 84         }

 85         res += (j > i ? j - i : 0);

 86     }

 87     return res;

 88 }

 89 int solve(int root)

 90 {

 91     int ans = 0;

 92     Get_size(root, -1);

 93     int s = FindGravity(root, -1, siz[root]).second;

 94     if (s == -1)

 95         return 0;

 96     vis[s] = true;

 97     for (int i = 0; i < G[s].size(); i++)

 98     {

 99         int v = G[s][i].to;

100         if (vis[v] == true)

101             continue;

102         ans += solve(v);

103     }

104     vector<int>ds;

105     ds.push_back(0);

106     for (int i = 0; i < G[s].size(); i++)

107     {

108         int v = G[s][i].to;

109         if (vis[v] == true)

110             continue;

111         vector<int>rds;

112         Get_len(v, s, G[s][i].len, rds);

113         ans -= cnt_pair(rds);

114         ds.insert(ds.end(), rds.begin(), rds.end());

115     }

116     ans += cnt_pair(ds);

117     vis[s] = false;

118     return ans;

119 }

120 int main()

121 {

122     #ifndef ONLINE_JUDGE

123         freopen("in.txt","r",stdin);

124     #endif

125     int n;

126     while ( scanf ("%d%d", &n, &K), n && K)

127     {

128         int u, v, c;

129         for (int i = 0; i <= n; i++)

130             G[i].clear();

131         for (int i = 0; i < n-1; i++)

132         {

133             scanf ("%d%d%d", &u, &v, &c);

134             G[u].push_back(Edge(v,c));

135             G[v].push_back(Edge(u,c));

136         }

137         printf("%d\n", solve(n/2+1));

138     }

139     return 0;

140 }

 ②第二种姿势，，200ms左右

  1 #include <set>

  2 #include <map>

  3 #include <cmath>

  4 #include <ctime>

  5 #include <queue>

  6 #include <stack>

  7 #include <cstdio>

  8 #include <string>

  9 #include <vector>

 10 #include <cstdlib>

 11 #include <cstring>

 12 #include <iostream>

 13 #include <algorithm>

 14 using namespace std;

 15 typedef unsigned long long ull;

 16 typedef long long ll;

 17 const int inf = 0x3f3f3f3f;

 18 const double eps = 1e-8;

 19 const int maxn = 1e4+10;

 20 struct Edge

 21 {

 22     int to, len, next;

 23 }e[maxn << 1];

 24 int head[maxn], tot, N, K;

 25 void add_edge(int u, int v, int c)

 26 {

 27     e[tot].to = v;

 28     e[tot].len = c;

 29     e[tot].next = head[u];

 30     head[u] = tot++;

 31 }

 32 int siz[maxn],Mtree[maxn];  // Mtree为最大子树的大小 siz为子树的大小

 33 bool vis[maxn];

 34 int center;

 35 void FindGravity(int u, int father, int cnt)  // 查找重心 center

 36 {

 37     siz[u] = 1;

 38     Mtree[u] = 0;

 39     for (int i = head[u]; ~i; i = e[i].next)

 40     {

 41         int v = e[i].to;

 42         if (v == father || vis[v] == true)

 43             continue;

 44         FindGravity(v, u, cnt);

 45         siz[u] += siz[v];

 46         Mtree[u] = max(Mtree[u], siz[v]);

 47     }

 48     Mtree[u] = max(cnt - siz[u], Mtree[u]);

 49     if (Mtree[center] > Mtree[u])

 50         center = u;

 51 }

 52 int S[maxn], dep[maxn], top;

 53 void Get_len(int u, int father, int d)

 54 {

 55     S[top++] = d;

 56     for (int i = head[u]; ~i; i = e[i].next)

 57     {

 58         int v = e[i].to;

 59         if (vis[v] == true || v == father)

 60             continue;

 61       //  dep[v] = dep[u] + e[i].len;

 62         Get_len(v, u, d+e[i].len);

 63     }

 64 }

 65 int Get_cnt(int u, int d)

 66 {

 67     int res = 0;

 68     top = 0;

 69     //dep[u] = d;

 70     Get_len(u, 0, d);

 71     sort (S, S+top);

 72     int j = top - 1;

 73     for (int i = 0; i < top; i++)

 74     {

 75         while (j > i && S[i] + S[j] > K)

 76             j--;

 77         res += (j > i ? j-i : 0);

 78     }

 79     return res;

 80 }

 81 int ans;

 82 void solve(int r)

 83 {

 84     vis[r] = true;

 85     ans += Get_cnt(r, 0);

 86     for (int i = head[r]; ~i; i = e[i].next)

 87     {

 88         int v = e[i].to;

 89         if (vis[v] == true)

 90             continue;

 91         ans -= Get_cnt(v, e[i].len);

 92         center = 0;

 93         FindGravity(v, r, siz[v]);

 94         solve(center);

 95     }

 96 }

 97 void init()

 98 {

 99     tot = 0;

100     Mtree[0] = N;

101     memset(head, -1, sizeof(head));

102     memset(vis, false, sizeof(vis));

103     center = 0;

104 }

105 int main()

106 {

107     #ifndef ONLINE_JUDGE

108         freopen("in.txt","r",stdin);

109     #endif

110     while (scanf ("%d%d", &N,&K), N && K)

111     {

112         init();

113         int u, v, c;

114         for (int i = 0; i < N-1; i++)

115         {

116             scanf ("%d%d%d", &u, &v, &c);

117             add_edge(u, v, c);

118             add_edge(v, u, c);

119         }

120         ans = 0;

121         FindGravity(N/2+1, -1, N);

122         solve(center);

123         printf("%d\n", ans);

124     }

125     return 0;

126 }