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题目描述 Description
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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. |
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输入描述 Input Description
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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. |
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输出描述 Output Description
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For each test case output the answer on a single line.
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样例输入 Sample Input
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5 4 1 2 3 1 3 1 1 4 2 3 5 1 0 0 |
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样例输出 Sample Output
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8
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数据范围及提示 Data Size & Hint
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一道比较裸的点分治题。对于每一棵树，先找到他的重心，然后算出他的所有子孙到他的dis,统一放在dis数组中，下面只需在数组中找数对满足和为k就好了，这个能在O（n）的时间内解决。但是这样会有一个问题。对于在同一棵子树上的点，路径并不是从这个点跑到重心，再跑下来，于是我们需要去重。去重之后对重心打上标记，表示以后不能再用这个点了，之后递归的处理每一个子树。一道破题调了好久，后来发现是在getd函数中传参数时出了BUG，看来以后函数的参数还是要有顺序的！
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 ",ans); 99 } 100 return 0; 101 }