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ec02f92
Create 3592.Inverse-Coin-Change.cpp
wisdompeak 67c5ed5
Update Readme.md
wisdompeak e918891
Create Readme.md
wisdompeak 97f67fa
Create 3593.Minimum-Increments-to-Equalize-Leaf-Paths.cpp
wisdompeak 2f2a9ad
Update Readme.md
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37 changes: 37 additions & 0 deletions
...imum-Increments-to-Equalize-Leaf-Paths/3593.Minimum-Increments-to-Equalize-Leaf-Paths.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,37 @@ | ||
| class Solution { | ||
| vector<int>adj[100005]; | ||
| public: | ||
| pair<long long, int>dfs(int u, int p, vector<int>&cost) { | ||
| long long maxPath = 0; | ||
| int totalChanged = 0; | ||
| vector<long long>paths; | ||
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| for (int v: adj[u]) { | ||
| if (v==p) continue; | ||
| auto [path, changed] = dfs(v, u, cost); | ||
| maxPath = max(maxPath, path); | ||
| totalChanged += changed; | ||
| paths.push_back(path); | ||
| } | ||
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| int count = 0; | ||
| for (long long p: paths) { | ||
| if (p < maxPath) | ||
| count++; | ||
| } | ||
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| return {cost[u]+maxPath, totalChanged + count}; | ||
| } | ||
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| int minIncrease(int n, vector<vector<int>>& edges, vector<int>& cost) { | ||
| for (auto& edge: edges) { | ||
| int u = edge[0], v = edge[1]; | ||
| adj[u].push_back(v); | ||
| adj[v].push_back(u); | ||
| } | ||
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| auto [_, ret ] = dfs(0, -1, cost); | ||
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| return ret; | ||
| } | ||
| }; |
26 changes: 26 additions & 0 deletions
Dynamic_Programming/3592.Inverse-Coin-Change/3592.Inverse-Coin-Change.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,26 @@ | ||
| class Solution { | ||
| public: | ||
| vector<int> findCoins(vector<int>& numWays) { | ||
| int n = numWays.size(); | ||
| numWays.insert(numWays.begin(), 0); | ||
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| vector<int>rets; | ||
| vector<int>dp(n+1); | ||
| dp[0] = 1; | ||
| for (int c=1; c<=n; c++) { | ||
| if (numWays[c] == dp[c]) | ||
| continue; | ||
| rets.push_back(c); | ||
| for (int i=c; i<=n; i++) { | ||
| dp[i] += dp[i-c]; | ||
| } | ||
| } | ||
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| for (int i=1; i<=n; i++) { | ||
| if (dp[i]!=numWays[i]) | ||
| return {}; | ||
| } | ||
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| return rets; | ||
| } | ||
| }; |
26 changes: 26 additions & 0 deletions
Dynamic_Programming/3592.Inverse-Coin-Change/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,26 @@ | ||
| ### 3592.Inverse-Coin-Change | ||
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| 这是一道非常有意思的"反向重构"的DP题。 | ||
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| 首先,从题境上来看这类似一道经典的背包问题。给出一系列硬币面值coins可以无限重复使用,问有多少种构造方法能拼出面值n来?对于这个问题,我们有如下动态规划解法。特别注意最外层的循环是硬币面额 | ||
| ``` | ||
| for (int c: coins) | ||
| for (int i=c; i<=n; i++) | ||
| dp[i] += dp[i-c]; | ||
| ``` | ||
| 状态转移的思想就是,每轮引入一种新的硬币面额c:遍历所有面值i,考察最后一个硬币是否使用c。如果不使用,那么dp[i]依然是前一轮的数值;如果使用c,那么就取决于dp[i-c]。 | ||
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| 我们可以沿用同样的思路,来思考numWays(也就是dp)是怎么计算出来的 | ||
| ``` | ||
| for (int c: coins) { | ||
| if c 不存在 | ||
| continue; | ||
| else if c 存在 | ||
| for (int i=c; i<=n; i++) | ||
| dp[i] += dp[i-c]; | ||
| } | ||
| ``` | ||
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| 那么如何判断c是否存在呢?突破点就是面额c的构造方法。如果numWays[c]等于前一轮(还没有引入面额c的时候)的dp[c],那么就意味着面额c一定不存在。否则c的构造必然至少还能增加一种方法(即只使用一个c硬币)。反之,面额c必须存在,否则无法弥补这个差额。 | ||
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| 当然,我们仅凭numWays[c]的分析来决定面额的种类还是比较片面的,它不能保证最终据此计算得到的dp都与所有的numsWays一致。所以我们还要再次校验一下。 |
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