AtCoder Beginner Contest 352
A
easy
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| #include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
const int N=1e6+100;
#define LF(x) fixed<<setprecision(x)
int a[N];
signed main()
{
int x,y,z,k;
cin>>x>>y>>z>>k;
if(y>z)
{
swap(y,z);
}
if(k>=y&&k<=z)
{
cout<<"Yes"<<'\n';
}else
{
cout<<"No";
}
return 0;
}
|
B
双指针 easy
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| #include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
const int N=1e6+100;
#define LF(x) fixed<<setprecision(x)
int a[N];
signed main()
{
string s,x;
cin>>s>>x;
int n=s.length()-1;
int m=x.length()-1;
int i=0;
int j=0;
while(i<=n)
{
if(x[j]==s[i])
{
i++;
j++;
cout<<j<<" ";
}
else
{
j++;
}
}
}
|
C
O（n)扫一遍，easy
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| #include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
const int N=1e6+100;
#define LF(x) fixed<<setprecision(x)
int a[N];
int b[N];
int sum;
signed main()
{
int n;
cin>>n;
rep(i,1,n)
{
cin>>a[i]>>b[i];
sum+=a[i];
}
int ans=0;
rep(i,1,n)
{
ans=max(sum-a[i]+b[i],ans);
}
cout<<ans;
}
|
D
注意到可以乱序，可以记录每个数字的下标，考虑用set维护，直接O(n)扫一遍
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| #include <bits/stdc++.h>
using namespace std;
int n, k;
int main(){
cin >> n >> k;
vector<int> q(n);
for(int i=0; i<n; i++){
int p;
cin >> p;
q[p-1] = i;
}
set<int> st;
for(int i=0; i<k; i++){
st.insert(q[i]);
}
int ans = *st.rbegin() - *st.begin();
for(int i=k; i<n; i++){
st.erase(q[i-k]);
st.insert(q[i]);
ans = min(ans, *st.rbegin() - *st.begin());
}
cout << ans << endl;
return 0;
}
|
E
板子题，不过存边需要一点注意，只需要存x-1条，不然会t，合理性自证。
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| #pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
int n,m,i,j,u,v,total;
const int N=1e7+100;
map<pair<int,int>,ll>mp;
struct edge
{
int start;
int to;
ll val;
}bian[N];
int f[N];
ll a[N];
long long ans;
int cnt_;
int find(int x)
{
if(f[x]==x)
{
return x;
}
else
{
return f[x]=find(f[x]);
}
}
bool cmp(edge a,edge b)
{
return a.val<b.val;
}
void kruskal()
{
for(int i=1;i<=cnt_;i++)
{
u=find(bian[i].start);
v=find(bian[i].to);
if(u==v)
{
continue;
}
ans+=bian[i].val;
f[u]=v;
total++;
if(total==n-1)
{
break;
}
}
}
signed main()
{
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
{
f[i]=i;
}
for(int i=1;i<=m;i++)
{
int x;
ll w;
scanf("%d %lld",&x,&w);
for(int j=1;j<=x;j++)
{
scanf("%d",&a[j]);
}
for(int j=2;j<=x;j++)
{
if(!mp[make_pair(a[j],a[1])])
{
bian[++cnt_].start=a[j];
bian[cnt_].to=a[1];
bian[cnt_].val=w;
mp[make_pair(a[1],a[1])]=cnt_;
}
else
{
bian[mp[make_pair(a[j],a[1])]].val=min(bian[mp[make_pair(a[j],a[1])]].val,w);
}
}
}
sort(bian+1,bian+cnt_+1,cmp);
kruskal();
if(total==n-1)
{
printf("%lld",ans);
}
else
{
printf("-1");
}
}
|
