題目鏈接

http://acm.split.hdu.edu.cn/showproblem.php?pid=5877

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Problem Description You are given a?rooted?tree of?N?nodes, labeled from 1 to?N. To the?ith node a non-negative value?ai?is assigned.An?ordered?pair of nodes?(u,v)?is said to be?weak?if
??(1)?u?is an ancestor of?v?(Note: In this problem a node?u?is not considered an ancestor of itself);
??(2)?au×av≤k.

Can you find the number of weak pairs in the tree? Input There are multiple cases in the data set.
??The first line of input contains an integer?T?denoting number of test cases.
??For each case, the first line contains two space-separated integers,?N?and?k, respectively.
??The second line contains?N?space-separated integers, denoting?a1?to?aN.
??Each of the subsequent lines contains two space-separated integers defining an edge connecting nodes?u?and?v?, where node?u?is the parent of node?v.

??Constrains:?
??1≤N≤105?
??0≤ai≤109?
??0≤k≤1018 Output For each test case, print a single integer on a single line denoting the number of weak pairs in the tree. Sample Input 1 2 3 1 2 1 2 Sample Output 1 題意：輸入n，k ?表示有n個節點的一棵樹，然后輸入n個節點的權值和n-1條邊，求點對（u，v）的對數，滿足： 1、u是v的祖先節點。 2、a[u]*a[v]<=k，a[]是存儲權值的數組。 思路：從根節點開始向下深搜，每到一個點時計算sum+=Sum（a[i]）,Sum(x)表示大于等于x的個數，然后向樹狀數組中加入k/a[i]，繼續遞歸深搜，退棧時從樹狀數組中減去k/a[i] ,這樣可以保證樹狀數組中存的一直是一條到根節點的路徑值。大題思路如上，這里要做一個離散化的處理，輸入的權值<=1e9 ?k<=1e18 ?而只有1e5個點，所以可以離散到2*1e5 后處理； 題解中提示用treap計算大于等于x的個數，這樣可以不需要進行離散化； 第一次自己做出深搜的題，挺高興的^_^ ?看樣子我對深搜有了一點認識了 代碼如下： #include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <map> using namespace std; const long long maxn=200003; long long root; long long sum,k; long long in[100005]; vector<long long>g[100005]; long long a[100005]; long long b[200005];long long c[200005]; map<long long,long long>q;long long Lowbit(long long t) {return t&(t^(t-1)); } void add(long long x,long long t) {while(x > 0){c[x]+=t;x -= Lowbit(x);} } long long Sum(long long li) {long long s=0;while(li<200005){s+=c[li];li=li+Lowbit(li);}return s; }void dfs(long long t) {long long n=g[t].size();for(long long i=0;i<n;i++){long long v=g[t][i];sum+=(long long)Sum(q[a[v]]);if(a[v]==0) add(maxn,1);else add(q[k/a[v]],1);dfs(v);if(a[v]==0) add(maxn,-1);else add(q[k/a[v]],-1);} }int main() {long long T,N;scanf("%lld",&T);while(T--){q.clear();memset(in,0,sizeof(in));memset(c,0,sizeof(c));memset(b,0,sizeof(b));scanf("%lld%lld",&N,&k);for(long long i=1;i<=N;i++){scanf("%lld",&a[i]);b[2*i-2]=a[i];if(a[i]!=0)b[2*i-1]=k/a[i];g[i].clear();}sort(b,b+2*N);long long tot=0,pre=-1;for(long long i=0;i<2*N;i++){if(b[i]!=pre){pre=b[i];q[pre]=++tot;}}for(long long i=0;i<N-1;i++){long long aa,bb;scanf("%lld%lld",&aa,&bb);g[aa].push_back(bb);in[bb]++;}for(long long i=1;i<=N;i++)if(in[i]==0) { root=i; break; }sum=0;if(a[root]==0) add(maxn,1);else add(q[k/a[root]],1);dfs(root);printf("%lld\n",sum);}return 0; }

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轉載于:https://www.cnblogs.com/chen9510/p/5860058.html

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