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Tree
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 Source |
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第一道点分治TAT
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e4 + 10;
const int inf = 1e8;
int siz[maxn],f[maxn],root,tot,k;
int head[maxn],e,dist[maxn],d[maxn];
bool vis[maxn];
struct Edge
{
int to,w,next;
}G[maxn<<1];
void AddEdge(int u,int v,int w)
{
G[++e] = (Edge) {v,w,head[u]};head[u] = e;
G[++e] = (Edge) {u,w,head[v]};head[v] = e;
}
void getroot(int u,int fa)
{
siz[u] = 1;f[u] = 0;
for(int eid = head[u]; eid ; eid = G[eid].next) {
int v = G[eid].to;
if(v==fa||vis[v])continue;
getroot(v,u);
siz[u] += siz[v];
f[u] = max(f[u],siz[v]);
}
f[u] = max(f[u],tot - siz[u]);
if(f[u] < f[root]) root = u;
}
void getdist(int u,int fa)
{
dist[++dist[0]] = d[u];
for(int eid = head[u]; eid ; eid = G[eid].next) {
int v = G[eid].to;
if(v==fa||vis[v])continue;
d[v] = d[u] + G[eid].w;
getdist(v,u);
}
}
int cacu(int u,int now)
{
dist[0] = 0;
d[u] = now;
getdist(u,0);
sort(dist+1,dist+1+dist[0]);
int l = 1,r = dist[0];
int num = 0;
while(l < r) {
if(dist[l] + dist[r] <= k) {
num += r - l;
l++;
}else r--;
}
return num;
}
int solve(int u)
{
int ans = cacu(u,0);vis[u] = 1;
for(int eid = head[u]; eid ; eid = G[eid].next) {
int v = G[eid].to;
if(vis[v]) continue;
ans -= cacu(v,G[eid].w);
root = 0;
tot = siz[v];
getroot(v,u);
ans += solve(root);
}
return ans;
}
int main(int argc, char const *argv[])
{
int n;
while(scanf("%d%d",&n,&k)==2) {
if(!n)return 0;
for(int i = 1; i <= n; i++)head[i] = 0;
for(int i = 1; i <= n; i++)vis[i] = 0;
e = 0;
for(int i = 1; i < n; i++) {
int u,v,w;scanf("%d%d%d",&u,&v,&w);
AddEdge(u,v,w);
}
f[root = 0] = inf;
tot = n;
getroot(1,-1);
int ans = solve(root);
printf("%d\n", ans);
}
return 0;
}原文地址：http://blog.csdn.net/acvcla/article/details/46583705
