题目描述
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.
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.
输入
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.
The last test case is followed by two zeros.
输出
For each test case output the answer on a single line.
样例输入
5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0
样例输出
8
题目大意
多组测试数据，每次输入n、m，和一棵n个点的有边权的树，问你满足x到y距离小于等于m的无序点对(x,y)的个数是多少。
题解
树的点分治模板题，第一次写
考虑到路径只有两种情况，一是经过根节点，二是不经过根节点。
如果不经过根节点，那么一定经过最小公共子树的根节点，可以转化为问题一的子问题。
于是考虑怎么递归解决问题一。
对于根节点进行一次dfs，求出deep，并将其从小到大排序。
避免重复，只需要求出其中deep[x]≤deep[y]且deep[x]+deep[y]≤m的个数。
用i表示左指针，j表示右指针，i从左向右遍历。
如果deep[i]+deep[j]≤m，则点对(i,t)(i<t≤j)都符合题意，将j-i加入答案中，并且i++；否则j--。
然而这样还会重复计算在同一棵子树中的点对，所以再进行下一步dfs之前需要减去重复部分。
但是这样做会TLE。为什么？因为树可能会退化，导致选择链头时时间复杂度极大。
于是每次不能固定选择root，而是以重心作为root去处理，这样能保证时间复杂度再O(nlog2n)以下。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 10010
using namespace std;
int m , head[N] , to[N << 1] , len[N << 1] , next[N << 1] , cnt , si[N] , deep[N] , root , vis[N] , f[N] , sn , d[N] , tot , ans;
void add(int x , int y , int z)
{
to[++cnt] = y , len[cnt] = z , next[cnt] = head[x] , head[x] = cnt;
}
void getroot(int x , int fa)
{
f[x] = 0 , si[x] = 1;
int i;
for(i = head[x] ; i ; i = next[i])
if(to[i] != fa && !vis[to[i]])
getroot(to[i] , x) , si[x] += si[to[i]] , f[x] = max(f[x] , si[to[i]]);
f[x] = max(f[x] , sn - si[x]);
if(f[root] > f[x]) root = x;
}
void getdeep(int x , int fa)
{
d[++tot] = deep[x];
int i;
for(i = head[x] ; i ; i = next[i])
if(to[i] != fa && !vis[to[i]])
deep[to[i]] = deep[x] + len[i] , getdeep(to[i] , x);
}
int calc(int x)
{
tot = 0 , getdeep(x , 0) , sort(d + 1 , d + tot + 1);
int i = 1 , j = tot , sum = 0;
while(i < j)
{
if(d[i] + d[j] <= m) sum += j - i , i ++ ;
else j -- ;
}
return sum;
}
void dfs(int x)
{
deep[x] = 0 , vis[x] = 1 , ans += calc(x);
int i;
for(i = head[x] ; i ; i = next[i])
if(!vis[to[i]])
deep[to[i]] = len[i] , ans -= calc(to[i]) , sn = si[to[i]] , root = 0 , getroot(to[i] , 0) , dfs(root);
}
int main()
{
int n , i , x , y , z;
while(scanf("%d%d" , &n , &m) && (n || m))
{
memset(head , 0 , sizeof(head));
memset(vis , 0 , sizeof(vis));
cnt = 0 , ans = 0;
for(i = 1 ; i < n ; i ++ )
scanf("%d%d%d" , &x , &y , &z) , add(x , y , z) , add(y , x , z);
f[0] = 0x7fffffff , sn = n;
root = 0 , getroot(1 , 0) , dfs(root);
printf("%d\n" , ans);
}
return 0;
}