[POJ1741]Tree

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題目描述 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.

輸入描述 Input Description

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.

輸出描述 Output Description

For each test case output the answer on a single line.

範例輸入 Sample Input

5 41 2 31 3 11 4 23 5 10 0

範例輸出 Sample Output

8

資料範圍及提示 Data Size & Hint

一道比較裸的點分治題。對於每一棵樹，先找到他的重心，然後算出他的所有子孫到他的dis,統一放在dis數組中，下面只需在數組中找數對滿足和為k就好了，這個能在O（n）的時間內解決。但是這樣會有一個問題。對於在同一棵子樹上的點，路徑並不是從這個點跑到重心，再跑下來，於是我們需要去重。去重之後對重心打上標記，表示以後不能再用這個點了，之後遞迴的處理每一個子樹。一道破題調了好久，後來發現是在getd函數中傳參數時出了BUG，看來以後函數的參數還是要有順序的！

  1 #include<iostream>  2 #include<cmath>  3 #include<algorithm>  4 #include<cstring>  5 #include<cstdio>  6 #include<queue>  7 using namespace std;  8 typedef long long LL;  9 #define Pi acos(-1.0) 10 #define mem(a,b) memset(a,b,sizeof(a)) 11 inline int read() 12 { 13     int x=0,f=1;char c=getchar(); 14     while(!isdigit(c)){if(c==‘-‘)f=-1;c=getchar();} 15     while(isdigit(c)){x=x*10+c-‘0‘;c=getchar();} 16     return x*f; 17 } 18 const int maxn=10010; 19 struct Edge 20 { 21     int u,v,w,next; 22     Edge() {} 23     Edge(int _1,int _2,int _3,int _4) : u(_1),v(_2),w(_3),next(_4) {} 24 }e[2*maxn]; 25 int first[maxn],n,k,a,b,c,ans,size[maxn],masize[maxn],now_size,root,dis[maxn],end; 26 bool vis[maxn];  27 void addEdge(int i,int a,int b,int c) 28 { 29     e[i]=Edge(a,b,c,first[a]); 30     first[a]=i; 31 } 32 void gets(int u,int pa) 33 { 34     size[u]=1; 35     masize[u]=0; 36     for(int i=first[u];i!=-1;i=e[i].next) 37         if(!vis[e[i].v] && e[i].v!=pa) 38         { 39             gets(e[i].v,u); 40             size[u]+=size[e[i].v]; 41             masize[u]=max(size[e[i].v],masize[u]); 42         } 43 } 44 void getr(int r,int u,int pa) 45 { 46     masize[u]=max(masize[u],size[r]-size[u]); 47     if(masize[u]<now_size)now_size=masize[u],root=u; 48     for(int i=first[u];i!=-1;i=e[i].next) 49         if(!vis[e[i].v] && e[i].v!=pa)getr(r,e[i].v,u); 50 } 51 void getd(int u,int pa,int d) 52 { 53     dis[end++]=d; 54     for(int i=first[u];i!=-1;i=e[i].next) 55         if(!vis[e[i].v] && e[i].v!=pa)getd(e[i].v,u,d+e[i].w); 56 } 57 int calc(int u,int d) 58 { 59     end=0; 60     getd(u,-1,d); 61     int ret=0,l=0,r=end-1; 62     sort(dis,dis+end);  63     while(l<r) 64     { 65         while(dis[l]+dis[r]>k && l<r)r--; 66         ret+=(r-l);l++; 67     } 68     return ret; 69 } 70 void dfs(int u) 71 { 72     now_size=n; 73     gets(u,-1); 74     getr(u,u,-1); 75     ans+=calc(root,0); 76     vis[root]=1; 77     for(int i=first[root];i!=-1;i=e[i].next) 78         if(!vis[e[i].v]) 79         { 80             ans-=calc(e[i].v,e[i].w); 81             dfs(e[i].v); 82         } 83     return; 84 } 85 void init(){mem(first,-1);mem(size,0);mem(masize,0);mem(vis,0);mem(dis,0);ans=0;} 86 int main() 87 { 88     while(scanf("%d%d",&n,&k)!=EOF) 89     { 90         if(n==0 && k==0)break; 91         init(); 92         for(int i=0;i<n-1;i++) 93         { 94             a=read();b=read();c=read(); 95             addEdge(i*2,a,b,c);addEdge(i*2+1,b,a,c); 96         } 97         dfs(1); 98         printf("%d\n",ans);     99     }100     return 0;101 }

 

[POJ1741]Tree