Tree
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 15548 | Accepted: 5054 |
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.
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.
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
【思路】
设i到当前root的距离为d[i]，i属于belong[i]->belong[i]为当前root的儿子且i在belong[i]为根的树中。设Sum{E}为满足条件E的点对数。
情况分为两种：
1) 经过根节点
2) 不经过根节点，在根节点的一颗子树中。
其中2)可以递归求解。
对于1）我们要求的是Sum{d[i]+d[j]<=k && belong[i]!=belong[j]}，即为
Sum{d[i]+d[j]<=k} - Sum{ d[i]+d[j]<=k && belong[i]==belong[j]}
前后两项都可以转化为求一个序列a中满足d[a[i]]+d[a[j]]<=k的点对数。
先将a按照d值排序，基于单调性，我们可以给出一个O(n)的统计方法。于是问题得到解决。
总的时间复杂度为O(nlog2n)
【代码】
1 #include<cstdio> 2 #include<vector> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 7 const int N = 10000+10; 8 const int INF = 1e9; 9 10 struct Edge{ 11 int u,v,w; 12 Edge(int u=0,int v=0,int w=0):u(u),v(v),w(w){}; 13 }; 14 int n,K,l1,l2,tl,ans; 15 int siz[N],d[N],list[N],f[N],can[N]; 16 vector<Edge> es; 17 vector<int> g[N]; 18 void adde(int u,int v,int w) { 19 es.push_back(Edge(u,v,w)); 20 int m=es.size(); g[u].push_back(m-1); 21 } 22 23 void init() { 24 ans=0; es.clear(); 25 memset(can,1,sizeof(can)); 26 for(int i=0;i<=n;i++) g[i].clear(); 27 } 28 void dfs1(int u,int fa) { 29 siz[u]=1; 30 list[++tl]=u; 31 for(int i=0;i<g[u].size();i++) { 32 int v=es[g[u][i]].v; 33 if(v!=fa && can[v]) { 34 dfs1(v,u); 35 f[v]=u; siz[u]+=siz[v]; 36 } 37 } 38 } 39 int getroot(int u,int fa) { //寻找u子树重心 40 int pos,mn=INF; 41 tl=0; 42 dfs1(u,fa); 43 for(int i=1;i<=tl;i++) { 44 int y=list[i],d=0; 45 for(int j=0;j<g[y].size();j++) { 46 int v=es[g[y][j]].v; 47 if(v!=f[y] && can[v]) d=max(d,siz[v]); 48 } 49 if(y!=u) d=max(d,siz[u]-siz[y]); //上方 50 if(d<mn) mn=d , pos=y; //使大子结点数最小 51 } 52 return pos; 53 } 54 void dfs2(int u,int fa,int dis) { 55 list[++l1]=u; d[u]=dis; 56 for(int i=0;i<g[u].size();i++) { 57 int v=es[g[u][i]].v; 58 if(v!=fa && can[v]) dfs2(v,u,dis+es[g[u][i]].w); 59 } 60 } 61 int getans(int* a,int l,int r) { 62 int res=0,j=r; 63 for(int i=l;i<=r;i++) { 64 while(d[a[i]]+d[a[j]]>K && j>i) j--; 65 res+=j-i; if(i==j) break; 66 } 67 return res; 68 } 69 bool cmp(const int& x,const int& y) { return d[x]<d[y]; } 70 void solve(int u,int fa) { 71 int root=getroot(u,fa); 72 l1=l2=0; 73 for(int i=0;i<g[root].size();i++) { //统计 d[i]+d[j]<=K && belong[i]==belong[j] 74 int v=es[g[root][i]].v; 75 if(can[v]) { 76 l2=l1; 77 dfs2(v,root,es[g[root][i]].w); //insert[以v为根的子树] 78 sort(list+l2+1,list+l1+1,cmp); 79 ans-=getans(list,l2+1,l1); 80 } 81 } 82 list[++l1]=root; d[root]=can[root]=0; 83 sort(list+1,list+l1+1,cmp); 84 ans+=getans(list,1,l1); //统计d[i]+d[j]<=K 85 for(int i=0;i<g[root].size();i++) { //递归<-分治 86 int v=es[g[root][i]].v; 87 if(v!=fa && can[v]) solve(v,root); 88 } 89 } 90 91 int main() { 92 while(scanf("%d%d",&n,&K)==2 && (n&&K)) { 93 int u,v,w; 94 init(); 95 for(int i=0;i<n-1;i++) { 96 scanf("%d%d%d",&u,&v,&w); 97 adde(u,v,w) , adde(v,u,w); 98 } 99 solve(1,-1); 100 printf("%d ",ans); 101 } 102 return 0; 103 }