Educational Codeforces Round 124 (Rated for Div. 2)

Dashboard - Educational Codeforces Round 124 (Rated for Div. 2) - Codeforceshttps://codeforces.com/contest/1651

A. Playoff

签到题，第一轮以后剩下的都是奇数，奇数又取最大，答案就是2的n次方-1

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'const int maxn = 0;
const int mod = 1e7+9;int main(){ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);int t;cin>>t;while(t--){int n;cin>>n;long long ans = pow(2,n)-1;cout<<ans<<endl;}
}

B. Prove Him Wrong

公式推导一下，如果YES，那么必然是x1+x2>=2(x2-x1)，也就是3x1>=x2，能尽可能让范围在数据内的情况是1 3 9....打个表n<20的时候最后的数在1e9以内

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'const int maxn = 0;
const int mod = 1e7+9;int main(){ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);int t;cin>>t;while(t--){int n;cin>>n;if(n>=20)cout<<"NO"<<endl;else{cout<<"YES"<<endl;int a = 1;for(int i = 1;i<=n;i++) {cout << a << " ";a *= 3;}cout<<endl;}}
}

C. Fault-tolerant Network

让首尾的度都变成2就可以了，由此推出七种情况(具体看代码吧)

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'const int maxn = 2e5+10;
const int mod = 1e7+9;
int a[maxn] = {};
int b[maxn] = {};int main(){ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);int t;cin>>t;while(t--){int n;cin>>n;for(int i = 1;i<=n;i++)cin>>a[i];for(int i = 1;i<=n;i++)cin>>b[i];long long ad1 = 1e9+10,ad2 = 1e9+10; //首尾到其他点的距离long long bd1 = 1e9+10,bd2 = 1e9+10; //首尾到其他点的距离long long abh = -1,abw = -1; //首首和尾尾long long abhw = -1,abwh = -1;//首尾和尾首abh = abs(a[1]-b[1]);abw = abs(a[n]-b[n]);abhw = abs(a[1]-b[n]);abwh = abs(b[1]-a[n]);for(int i = 1;i<=n;i++){long long d = abs(a[1]-b[i]);ad1 = min(ad1,d); //首到其他d = abs(a[n]-b[i]);ad2 = min(ad2,d);}for(int i = 1;i<=n;i++){long long d = abs(b[1]-a[i]);bd1 = min(bd1,d);d = abs(b[n]-a[i]);bd2 = min(bd2,d);}// cout<<ad1<<" "<<ad2<<" "<<bd1<<" "<<bd2<<" "<<abh<<" "<<abw<<" "<<abhw<<" "<<abwh<<endl;long long ans = min(abhw+abwh,abh+abw);ans = min(ans,abh+ad2+bd2);  //首首连ans = min(ans,abw+ad1+bd1);ans = min(ans,abhw+ad2+bd1);ans = min(ans,abwh+ad1+bd2);ans = min(ans,ad1+ad2+bd1+bd2);cout<<ans<<endl;}
}

D. Nearest Excluded Points

多源bfs，目的就是减少单点bfs的反复搜索，由贪心思路可知，两种情况，一种是身边有空的，那么答案就是空的，一种是身边没有空的，那么必然它的答案是身边的某一非空点的答案，再使用队列保证一旦找到就是最近距离之一即可。

#include <bits/stdc++.h>#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn = 2e5+10;
pair<int,int>p[maxn];
map<pair<int,int>,pair<int,int>>ans;
set<pair<int,int>>vis;
queue<pair<int,int>>q;
pair<int,int>now;
pair<int,int>nex;
int dx[] = {0,0,0,1,-1};
int dy[] = {0,1,-1,0,0};void bfs(){while(!q.empty()){now = q.front();q.pop();for(int i = 1;i<=4;i++){nex = {now.first+dx[i],now.second+dy[i]};if(vis.count(nex)&&!ans.count(nex)){ //被围死的点的答案和被围着的某个点一样ans[nex] = ans[now];q.push(nex);}}}
}int main() {int n;cin>>n;for(int i = 1;i<=n;i++)cin >> p[i].first >> p[i].second;for(int i = 1;i<=n;i++)vis.insert(p[i]);/*for(int i = 1;i<=n;i++)cout<<p[i].first<<" "<<p[i].second<<endl;*/for(int i = 1;i<=n;i++){for(int j = 1;j<=4;j++) {now = {p[i].first + dx[j], p[i].second + dy[j]};if(!vis.count(now)) {ans[p[i]] = now;q.push(p[i]);break;}}}bfs();for(int i = 1;i<=n;i++)cout<<ans[p[i]].first<<" "<<ans[p[i]].second<<endl;
}

加油

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