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110 changes: 110 additions & 0 deletions
..._Search/3585.Find-Weighted-Median-Node-in-Tree/3585.Find-Weighted-Median-Node-in-Tree.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,110 @@ | ||
| using ll = long long; | ||
| const int MAXN = 100000; | ||
| const int LOGN = 17; | ||
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| class Solution { | ||
| public: | ||
| vector<pair<int,int>> adj[MAXN]; | ||
| int up[MAXN][LOGN+1]; | ||
| int depth[MAXN]; | ||
| ll distRoot[MAXN]; | ||
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| void dfs(int cur, int parent) | ||
| { | ||
| up[cur][0] = parent; | ||
| for(auto &[v,w]: adj[cur]) | ||
| { | ||
| if(v == parent) continue; | ||
| depth[v] = depth[cur] + 1; | ||
| distRoot[v] = distRoot[cur] + w; | ||
| dfs(v, cur); | ||
| } | ||
| } | ||
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| int lca(int a, int b) | ||
| { | ||
| if(depth[a] < depth[b]) swap(a,b); | ||
| int diff = depth[a] - depth[b]; | ||
| for(int k = 0; k <= LOGN; k++){ | ||
| if(diff & (1<<k)) a = up[a][k]; | ||
| } | ||
| if(a == b) return a; | ||
| for(int k = LOGN; k >= 0; k--){ | ||
| if(up[a][k] != up[b][k]){ | ||
| a = up[a][k]; | ||
| b = up[b][k]; | ||
| } | ||
| } | ||
| return up[a][0]; | ||
| } | ||
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| int stepUp(int u, int k) { | ||
| for (int i=16; i>=0; i--) { | ||
| if ((k>>i)&1) { | ||
| u = up[u][i]; | ||
| } | ||
| } | ||
| return u; | ||
| } | ||
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| ll dist(int a, int b) | ||
| { | ||
| int c = lca(a,b); | ||
| return distRoot[a] + distRoot[b] - 2*distRoot[c]; | ||
| } | ||
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| vector<int> findMedian(int n, vector<vector<int>>& edges, vector<vector<int>>& queries) { | ||
| for (int i = 0; i < n-1; i++) | ||
| { | ||
| int u = edges[i][0], v = edges[i][1], w = edges[i][2]; | ||
| adj[u].push_back({v,w}); | ||
| adj[v].push_back({u,w}); | ||
| } | ||
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| depth[0] = 0; | ||
| distRoot[0] = 0; | ||
| dfs(0, 0); | ||
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| for(int k = 1; k <= LOGN; k++) { | ||
| for(int v = 0; v < n; v++) { | ||
| up[v][k] = up[up[v][k-1]][k-1]; | ||
| } | ||
| } | ||
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| vector<int>rets; | ||
| for (auto& q: queries) | ||
| { | ||
| int u = q[0], v = q[1]; | ||
| int c = lca(u,v); | ||
| ll total = dist(u,c)+dist(c,v); | ||
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| int step1 = depth[u]-depth[c]; | ||
| int step2 = depth[v]-depth[c]; | ||
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| int low = 0, high = step1+step2; | ||
| int k; | ||
| while (low < high) { | ||
| int mid = low + (high-low)/2; | ||
| ll d; | ||
| if (mid <= step1) { | ||
| k = stepUp(u, mid); | ||
| d = distRoot[u] - distRoot[k]; | ||
| } else { | ||
| k = stepUp(v, step2 - (mid-step1)); | ||
| d = total - (distRoot[v] - distRoot[k]); | ||
| } | ||
| if (d >= total*0.5) | ||
| high = mid; | ||
| else | ||
| low = mid+1; | ||
| } | ||
| int step = low; | ||
| if (step<=step1) | ||
| rets.push_back(stepUp(u, step)); | ||
| else | ||
| rets.push_back(stepUp(v, step2-(step-step1))); | ||
| } | ||
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| return rets; | ||
| } | ||
| }; |
7 changes: 7 additions & 0 deletions
Binary_Search/3585.Find-Weighted-Median-Node-in-Tree/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,7 @@ | ||
| ### 3585.Find-Weighted-Median-Node-in-Tree | ||
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| 对于任何一个query,我们只需要找到u到v路径(途中经过LCA的点记作c),假设路径的总步长是d,路径的总权重和是total。我们只需要在[0,d]之间进行二分搜索一个合适的步数k:即从u走k步,恰好走过的路径长度超过total的一半。 | ||
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| 注意,我们在二分搜索对k进行判定的时候,需要分类讨论k是否在u到c的路径上,还是c到v的路径上。即看是否`dist(u,c) >= total * 0.5`. 如果k是在u到c的路径上,那么经过的路径长度就是dist(u,k)。如果k是在c到v的路径上,那么经过的路径长度就是dist(u,c)+dist(c,k)。 | ||
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| 根据binary lifting的算法,树里任意两个节点之间的距离都可以用log(n)的时间求解。 |
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