https://atcoder.jp/contests/abc395
https://www.bilibili.com/video/BV1vE98YDEXA/?vd_source=6759ab7a746b50893c564c06fbc6a752
C - Shortest Duplicate Subarray
https://atcoder.jp/contests/abc395/tasks/abc395_c
这一题很简单
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| #include<bits/stdc++.h> using namespace std; using u32 = unsigned; #define i128 __int128; using ll = long long;
using u64 = unsigned long long; const ll inf = 1e9; const ll INF = 1e18;
signed main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n; cin>>n; map<int,int>mp; int ans=INT_MAX; for(int i=1;i<=n;i++) { int x; cin>>x; if(mp.contains(x)) { ans=min(ans,i-mp[x]+1); mp[x]=i; } else { mp[x]=i; } } cout<<(ans==INT_MAX?-1:ans)<<'\n'; return 0; }
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D - Pigeon Swap
https://atcoder.jp/contests/abc395/tasks/abc395_d
这一题用了偷懒的方法,在步骤二中直接切换两个鸽子笼的编号。
平常没做过这种题,即使想出来了在实现的时候也遇到了很多困难,把自己绕进去好几次。
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| #include<bits/stdc++.h> using namespace std; using u32 = unsigned; #define i128 __int128; using ll = long long;
using u64 = unsigned long long; const ll inf = 1e9; const ll INF = 1e18;
signed main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n; cin>>n; map<int,int>mp; int ans=INT_MAX; for(int i=1;i<=n;i++) { int x; cin>>x; if(mp.contains(x)) { ans=min(ans,i-mp[x]+1); mp[x]=i; } else { mp[x]=i; } } cout<<(ans==INT_MAX?-1:ans)<<'\n'; return 0; }
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E - Flip Edge
https://atcoder.jp/contests/abc395/tasks/abc395_e
非常标准的双层图,建图之后直接跑dijkstra就行了。
但是一般都是多开一维数组,根据当前状态去走。不然层数多状态多的话根本无法存下所有的边。
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| #include<bits/stdc++.h> using namespace std; using u32 = unsigned; #define i128 __int128; using ll = long long; #define int ll using u64 = unsigned long long; const ll inf = 1e9; const ll INF = 1e18; int n,m,x; const int N=2e5+10; vector<pair<int,int>>adj[2*N]; int dis[2*N]; bool vis[2*N]; signed main() { ios::sync_with_stdio(false); cin.tie(nullptr); cin>>n>>m>>x; for(int i=1;i<=m;i++) { int u,v; cin>>u>>v; adj[u].push_back({v,1}); adj[v+n].push_back({u+n,1}); } for(int i=1;i<=n;i++) { adj[i].push_back({i+n,x}); adj[i+n].push_back({i,x}); } fill(begin(dis),end(dis),LLONG_MAX); dis[1]=0; priority_queue<pair<int,int>,vector<pair<int,int>>,greater<>>pq; pq.push({0,1}); while(pq.size()) { auto [_,u]=pq.top(); pq.pop(); if(vis[u])continue; vis[u]=1; for(auto [v,w]:adj[u]) { if(dis[u]+w<dis[v]) { dis[v]=dis[u]+w; pq.push({dis[v],v}); } } } cout<<min(dis[n],dis[2*n])<<'\n';
return 0; }
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