題干：

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

題目大意：

第一行給定n和k，然后給定一棵n個點的無向樹，每條邊有邊權v，求點對(u,v)的個數，使得dis[u][v]<=k，n<=1e4

解題報告：

? 點分治裸題，注意求重心的時候別大于號小于號別弄反了，應該是求最大子樹最小的頂點。還有別忘size的重構，很多代碼貌似換根之后都沒有重構，這樣得到的all變量應該就是不對的，相應求出的當前子樹的重心也應該是不對的。哦對了，別忘容斥。

AC代碼：

//嚴格樹的重心 #include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int tot,head[MAX]; int n,k; ll ans ; struct Edge {int u,v,w,ne; } e[MAX]; void add(int u,int v,int w) {e[++tot].u = u; e[tot].v = v; e[tot].w = w;e[tot].ne = head[u]; head[u] = tot; } int size[MAX],vis[MAX],son[MAX],rt,all; void getRoot(int cur,int fa) {size[cur] = 1; son[cur] = 0;for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(v == fa || vis[v]) continue;getRoot(v,cur);size[cur] += size[v];son[cur] = max(son[cur],size[v]); }son[cur] = max(son[cur],all - size[cur]);if(son[rt] == 0 || son[rt] > son[cur]) rt = cur; } int tott,dis[MAX]; void getdis(int cur,int fa,int ddis) {dis[++tott] = ddis;for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(v == fa || vis[v]) continue;getdis(v,cur,ddis + e[i].w); } } int cal(int cur,int diss) {tott = 0; getdis(cur,0,diss);sort(dis+1,dis+tott+1);int l = 1, r = tott,res = 0;for(;l<r; l++) {while(dis[l]+dis[r] > k && r > l) r--;res += r-l;}return res; } void gx(int cur,int fa) {size[cur] = 1;for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(v == fa || vis[v]) continue;gx(v,cur);size[cur] += size[v]; } } void dfs(int cur) {rt = 0;getRoot(cur,0);cur = rt;//getRoot(cur,0);vis[cur] = 1;gx(cur,0);ans += cal(cur,0);for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(vis[v]) continue;ans -= cal(v,e[i].w);all = size[v]; dfs(v);} } void init() {for(int i = 1; i<=n; i++) {vis[i] = size[i] = 0;head[i] = -1; }tot=ans=0; } int main() {while(~scanf("%d%d",&n,&k) && n+k) {init(); for(int u,v,w,i = 1; i<=n-1; i++) {scanf("%d%d%d",&u,&v,&w);add(u,v,w);add(v,u,w);} dfs(1);printf("%lld\n",ans);} return 0 ; }

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