AtCoder Beginner Contest 377
D - Many Segments 2
https://atcoder.jp/contests/abc377/tasks/abc377_d
创建一个dp数组，dp[i]为第i个位置右边第一个区间的右边界。题中给了n组l和r，对于l来说,dp[l]=r;对于其他位置来说，从右往左传递状态，dp[i]=min(dp[i],dp[i+1])，然后对于每一个位置l来说，r所能到达的最右边的位置就是dp[l]-1，所以对于每一个l来说，有dp[l]-l个区间。
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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 = 2147483647;
const ll INF = 1e18;
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,m;
cin>>n>>m;
vector<int>dp(m+1,m+1);
for(int i=1;i<=n;i++)
{
int l,r;
cin>>l>>r;
dp[l]=min(dp[l],r);
}
for(int i=m-1;i>=1;i--)
{
dp[i]=min(dp[i],dp[i+1]);
}
ll cnt=0;
for(int i=1;i<=m;i++)
{
cnt+=(dp[i]-i);
}
cout<<cnt<<'\n';
return 0;
}
|
E - Permute K times 2
https://atcoder.jp/contests/abc377/tasks/abc377_e
原始置换是A,
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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 = 2147483647;
const ll INF = 1e18;
ll power(ll x,ll y,int mod)
{
ll ret=1;
while(y)
{
if(y&1)ret=ret*x%mod;
x=x*x%mod;
y>>=1;
}
return ret;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
ll k;
cin>>n>>k;
vector<int>p(n+1);
for(int i=1;i<=n;i++)cin>>p[i];
vector<bool>vis(n+1);
for(int i=1;i<=n;i++)
{
if(vis[i])continue;
vector<int>a;
int j=i;
while(!vis[j])
{
vis[j]=1;
a.push_back(j);
j=p[j];
}
int d=power(2,k,a.size());
for(int x=0;x<a.size();x++)
{
p[a[x]]=a[(x+d)%a.size()];
}
}
for(int i=1;i<=n;i++)cout<<p[i]<<" \n"[i==n];
return 0;
}
|
F - Avoid Queen Attack
https://atcoder.jp/contests/abc377/tasks/abc377_f
这一题主要是有四部分，横（Row)，竖（Line)，对角线（D)，反对角线（E)
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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,m;
cin>>n>>m;
ll ans=1ll*n*n;
set<int>R,C,D,E;
for(int i=1;i<=m;i++)
{
int a,b;
cin>>a>>b;
R.insert(a);
C.insert(b);
D.insert(a-b);
E.insert(a+b);
}
ans-=1ll*R.size()*n;
ans-=1ll*C.size()*n;
for(auto d:D)
{
ans-=n-abs(d);
}
for(auto e:E)
{
ans-=n-abs(n+1-e);
}
for(auto r:R)然后分别依次减掉与其他部分重合的地方
{
set<array<int,2>>s;
for(auto c:C)
{
s.insert({r,c});
}
for(auto d:D)
{
int c=r-d;
if(1<=c&&c<=n)
{
s.insert({r,c});
}
}
for(auto e:E)
{
int c=e-r;
if(1<=c&&c<=n)
{
s.insert({r,c});
}
}
ans+=s.size();
}
for(auto c:C)
{
set<array<int,2>>s;
for(auto d:D)
{
int r=c+d;
if(1<=r&&r<=n)
{
s.insert({r,c});
}
}
for(auto e:E)
{
int r=e-c;
if(1<=r&&r<=n)
{
s.insert({r,c});
}
}
ans+=s.size();
}
for(auto d:D)
{
set<array<int,2>>s;
for(auto e:E)
{
if((1ll*d+e)%2)continue;
int r=(1ll*d+e)/2;
int c=(1ll*e-d)/2;
if(1<=r&&r<=n&&1<=c&&c<=n)s.insert({r,c});
}
ans+=s.size();
}
cout<<ans<<'\n';
return 0;
}
|
Edit to Match
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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;
const int N=2e5+10;
int Trie[N][26];
int dix;
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin>>n;
vector<int>dis(N+1,inf);
for(int i=1;i<=n;i++)
{
string s;
cin>>s;
int p=0;
int ans=s.size();
for(int j=0;j<s.size();j++)
{
int &q=Trie[p][s[j]-'a'];
if(!q)q=++dix;
p=q;
ans=min(ans,(int)s.size()-1-j+dis[p]);
}
p=0;
for(int j=0;j<s.size();j++)
{
p=Trie[p][s[j]-'a'];
dis[p]=min(dis[p],(int)s.size()-1-j);
}
cout<<ans<<'\n';
}
return 0;
}
|
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