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;
//#define int ll
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;
}

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;
//#define int ll
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;
}

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;
}