POJ 1741 Tree（树的分治）

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.

 

题目大意：给一个带边权的树，问有多少对点满足dist(i,j)<=k

思路：可以参考2009年的国家集训队论文《分治算法在树的路径问题中的应用》——漆子超。

 

代码（188MS）：

  1 #include <cstdio>

  2 #include <iostream>

  3 #include <algorithm>

  4 #include <cstring>

  5 using namespace std;

  6 

  7 const int MAXN = 10010;

  8 const int MAXE = 20010;

  9 const int INF = 0x7fff7fff;

 10 

 11 int head[MAXN], size[MAXN], maxSize[MAXN];

 12 int list[MAXN], cnt;

 13 bool del[MAXN];

 14 int to[MAXE], next[MAXE], cost[MAXE];

 15 int n, k, ecnt;

 16 

 17 void init() {

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

 19     memset(del, 0, sizeof(del));

 20     ecnt = 0;

 21 }

 22 

 23 void add_edge(int u, int v, int c) {

 24     to[ecnt] = v; cost[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;

 25     to[ecnt] = u; cost[ecnt] = c; next[ecnt] = head[v]; head[v] = ecnt++;

 26 }

 27 

 28 void dfs1(int u, int f) {

 29     size[u] = 1;

 30     maxSize[u] = 0;

 31     for(int p = head[u]; ~p; p = next[p]) {

 32         int &v = to[p];

 33         if(v == f || del[v]) continue;

 34         dfs1(v, u);

 35         size[u] += size[v];

 36         maxSize[u] = max(maxSize[u], size[v]);

 37     }

 38     list[cnt++] = u;

 39 }

 40 

 41 int get_root(int u, int f) {

 42     cnt = 0;

 43     dfs1(u, f);

 44     int ret, maxr = INF;

 45     for(int i = 0; i < cnt; ++i) {

 46         int &x = list[i];

 47         if(max(maxSize[x], size[u] - size[x]) < maxr) {

 48             ret = x;

 49             maxr = max(maxSize[x], size[u] - size[x]);

 50         }

 51     }

 52     return ret;

 53 }

 54 

 55 void dfs2(int u, int f, int dis) {

 56     list[cnt++] = dis;

 57     for(int p = head[u]; ~p; p = next[p]) {

 58         int &v = to[p];

 59         if(v == f || del[v]) continue;

 60         dfs2(v, u, dis + cost[p]);

 61     }

 62 }

 63 

 64 int calc(int a, int b) {

 65     int j = b - 1, ret = 0;

 66     for(int i = a; i < b; ++i) {

 67         while(list[i] + list[j] > k && i < j) --j;

 68         ret += j - i;

 69         if(j == i) break;

 70     }

 71     return ret;

 72 }

 73 

 74 int ans = 0;

 75 

 76 void work(int u, int f) {

 77     int root = get_root(u, f);

 78     del[root] = true;

 79     int last = 0; cnt = 0;

 80     for(int p = head[root]; ~p; p = next[p]) {

 81         int &v = to[p];

 82         if(del[v]) continue;

 83         dfs2(v, root, cost[p]);

 84         sort(list + last, list + cnt);

 85         ans -= calc(last, cnt);

 86         last = cnt;

 87     }

 88     list[cnt++] = 0;

 89     sort(list, list + cnt);

 90     ans += calc(0, cnt);

 91     for(int p = head[root]; ~p; p = next[p]) {

 92         int &v = to[p];

 93         if(del[v]) continue;

 94         work(v, root);

 95     }

 96 }

 97 

 98 int main() {

 99     while(scanf("%d%d", &n, &k) != EOF) {

100         if(n == 0 && k == 0) break;

101         init();

102         for(int i = 1; i < n; ++i) {

103             int u, v, c;

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

105             add_edge(u, v, c);

106         }

107         ans = 0;

108         work(1, 0);

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

110     }

111 }

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