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5290ed5
Create 3600.Maximize-Spanning-Tree-Stability-with-Upgrades.cpp
wisdompeak eed8742
Update Readme.md
wisdompeak 2ad5af3
Create Readme.md
wisdompeak c7d5507
Create union_find.cpp
wisdompeak da08f58
Update Readme.md
wisdompeak 461961a
Update 3600.Maximize-Spanning-Tree-Stability-with-Upgrades.cpp
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23 changes: 23 additions & 0 deletions
Template/Union_Find/union_find.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,23 @@ | ||
| struct DSU { | ||
| vector<int> p, r; | ||
| DSU(int n): p(n,-1), r(n,0) { | ||
| iota(p.begin(), p.end(), 0); | ||
| } | ||
| int find(int x) { | ||
| return p[x]==x ? x : p[x]=find(p[x]); | ||
| } | ||
| bool unite(int a, int b) { | ||
| a = find(a); b = find(b); | ||
| if (a == b) return false; | ||
| if (r[a] < r[b]) swap(a,b); | ||
| p[b] = a; | ||
| if (r[a]==r[b]) ++r[a]; | ||
| return true; | ||
| } | ||
| }; | ||
|
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| int main { | ||
| DSU dsu(n); | ||
| dsu.unite(u,v); | ||
| if (dsu.find(u)==dsu.find(v) {} | ||
| } |
75 changes: 75 additions & 0 deletions
...ning-Tree-Stability-with-Upgrades/3600.Maximize-Spanning-Tree-Stability-with-Upgrades.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,75 @@ | ||
| class Solution { | ||
| public: | ||
| struct DSU { | ||
| vector<int> p, r; | ||
| DSU(int n): p(n,-1), r(n,0) { | ||
| iota(p.begin(), p.end(), 0); | ||
| } | ||
| int find(int x) { | ||
| return p[x]==x ? x : p[x]=find(p[x]); | ||
| } | ||
| bool unite(int a, int b) { | ||
| a = find(a); b = find(b); | ||
| if (a == b) return false; | ||
| if (r[a] < r[b]) swap(a,b); | ||
| p[b] = a; | ||
| if (r[a]==r[b]) ++r[a]; | ||
| return true; | ||
| } | ||
| }; | ||
|
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| bool isOK(int T, int n, vector<vector<int>>& edges, int k) { | ||
| DSU dsu(n); | ||
| int count = 0; | ||
| int upgrade = 0; | ||
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| vector<vector<int>>candidates; // upgrade, u,v | ||
| for (auto& e: edges) { | ||
| int u = e[0], v = e[1], s = e[2], must = e[3]; | ||
| if (must) { | ||
| if (s<T) return false; | ||
| dsu.unite(u,v); | ||
| count++; | ||
| } else { | ||
| if (s>=T) | ||
| candidates.push_back({0, u, v}); | ||
| else if (2*s>=T) | ||
| candidates.push_back({1, u, v}); | ||
| } | ||
| } | ||
|
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| sort(candidates.begin(), candidates.end()); | ||
|
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| for (auto& cand: candidates) { | ||
| int u = cand[1], v = cand[2]; | ||
| if (count==n-1) break; | ||
| if (dsu.find(u)!=dsu.find(v)) { | ||
| dsu.unite(u,v); | ||
| count++; | ||
| upgrade += cand[0]; | ||
| if (upgrade > k) return false; | ||
| } | ||
| } | ||
|
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| for (int i=0; i<n; i++) | ||
| if (dsu.find(i)!=dsu.find(0)) | ||
| return false; | ||
|
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| return count==n-1; | ||
| } | ||
|
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| int maxStability(int n, vector<vector<int>>& edges, int k) { | ||
| int lo = 1, hi = 1e5*2; | ||
| while (lo < hi) { | ||
| int mid = hi-(hi-lo)/2; | ||
| if (isOK(mid, n, edges, k)) | ||
| lo = mid; | ||
| else | ||
| hi = mid-1; | ||
| } | ||
| if (isOK(lo, n, edges, k)) | ||
| return lo; | ||
| else | ||
| return -1; | ||
| } | ||
| }; |
7 changes: 7 additions & 0 deletions
Union_Find/3600.Maximize-Spanning-Tree-Stability-with-Upgrades/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,7 @@ | ||
| ### 3600.Maximize-Spanning-Tree-Stability-with-Upgrades | ||
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| 突破口是二分搜值。如果猜测全局最小的stability是T的话,尝试能否用最小生成树的方法构造出一棵树来。 | ||
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| 当固定全局最小的stability是T之后,那么哪些edge能用哪些不能用就一目了然了。对于must的那些边,如果存在stability小于T的,那么显然无解。对于非must的边,如果`2*stability<T`,那么必然排除作为构造树的候选;反之就可以作为候选,同时我们需要根据2s和T的比较,标注这条边"是否需要upgrade"。 | ||
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| 有了所有的候选边,我们就可以用Kruskal算法构造"最小生成树"。只不过这里优化的不是总边的权重,而是"upgrade的次数"。我们将所有边按照"是否需要upgrade"排序即可(其实就是0或者1),优先选用不需要upgrade的边实现更多的union。如果发现一个候选边的两个端点已经union在一起了,那么这条边其实就可以放心地舍弃而没有任何副作用。注意处理完所有的边之后还需要判断两点:是否恰好用到了n-1条边(否则说明可能有环),是否所有的点的find都相同(否则说明可能有不联通的区域)。 |
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