POJ 1741 Tree 解题报告（树分治）

Tree

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 10589 Accepted: 3257

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 41 2 31 3 11 4 23 5 10 0

Sample Output

8

Source

LouTiancheng@POJ

    解题报告：楼教主男人八题之一。树分治。

    我觉得难点在于，在你计算复杂度之前，你肯定不会想到这么做。

    每次同此当前每个节点子树的节点数，找到重心。这个复杂度为O(n)。统计重心到所有节点的距离，找到所有长度小于等于K的链。这里可以先排序，用O(n)的算法找到所有解。排序复杂度O(n log n)。删除重心，去重，递归下去。每次找的都是重心，可以保证递归的深度不超过log n。故总复杂度为O(n log n log n)。

    问题就解决啦……代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <iomanip>#include <cassert>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000")#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define travel(e, u) for(int e = u, v = vv[u]; e; e = nxt[e], v = vv[e])#define bit(n) (1LL<<(n))typedef long long LL;typedef unsigned long long ULL;void work();int main(){#ifdef ACM    freopen("in.txt", "r", stdin);#endif // ACM    work();}void scanf(int & x, char ch = 0){    while((ch = getchar()) < '0' || ch > '9');    x = ch - '0';    while((ch = getchar()) >= '0' && ch <= '9') x = 10 * x + (ch - '0');}/***************************************************************************************/const int maxv = 11111;int n, k;int ans;int edge[maxv], ecnt;int nxt[maxv * 2], vv[maxv * 2], ww[maxv * 2];bool vis[maxv];int siz[maxv], mson[maxv];int mi, root;int tot, dis[maxv];void init(){    ans = 0;    ecnt = 2;    memset(edge, 0, sizeof(edge));    memset(vis, 0, sizeof(vis));}void addEdge(int u, int v, int w, int first[]){    nxt[ecnt] = first[u], vv[ecnt] = v, ww[ecnt] = w, first[u] = ecnt++;}void dfsSize(int u, int f){    siz[u] = 1;    mson[u] = 0;    travel(e, edge[u]) if(!vis[v] && v != f)    {        dfsSize(v, u);        siz[u] += siz[v];        mson[u] = max(mson[u], siz[v]);    }}void dfsGravity(int r, int u, int f){    mson[u] = max(mson[u], siz[r] - siz[u]);    if(mson[u] < mi) mi = mson[u], root = u;    travel(e, edge[u]) if(!vis[v] && v != f)        dfsGravity(r, v, u);}void dfsDis(int u, int f, int d){    dis[tot++] = d;    travel(e, edge[u]) if(!vis[v] && v != f)        dfsDis(v, u, d + ww[e]);}int calc(int u, int d = 0){    tot = 0;    dfsDis(u, 0, d);    sort(dis, dis + tot);    int ret = 0;    int l = 0, r = tot - 1;    while(l < r)    {        while(dis[l] + dis[r] > k && l < r) r--;        ret += r - l;        l++;    }    return ret;}void dfs(int u){    mi = n;    dfsSize(u, 0);    dfsGravity(u, u, 0);    ans += calc(root);    vis[root] = true;    travel(e, edge[root]) if(!vis[v])    {        ans -= calc(v, ww[e]);        dfs(v);    }}void work(){    while(scanf("%d%d", &n, &k) == 2 && (n || k))    {        init();        ff(i, n-1)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            addEdge(u, v, w, edge);            addEdge(v, u, w, edge);        }        dfs(1);        cout << ans << endl;    }}