Educational Codeforces Round 105 (Rated for Div. 2) A-D题解

A. ABC String
题意给你一个只包含 A、B、C的字符串，每类字符都有可能代表左右括号中的一种，问你是否能找到一个方式使得括号匹配合法
三个字符，每个字符有两种情况总计 2^3 = 8次枚举 + 判断括号匹配合法性即可

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
using namespace std;#define ll long long
#define fi first
#define se second
#define pr pair<int,int>const int MAX_N = 55;
char str[MAX_N];int tran(int x)
{if(x == 1) return 1;return -1;
}int main()
{int t;scanf("%d",&t);while(t--){scanf("%s",str+1);int len = strlen(str+1);bool flag = false;for(int i = 1;i<=2;++i){for(int j = 1;j<=2;++j){for(int k = 1;k<=2;++k){bool flag_ = true;int cnt = 0;for(int l = 1;l<=len;++l){if(str[l]=='A'){cnt+=tran(i);}else if(str[l]=='B'){cnt+=tran(j);}else if(str[l]=='C'){cnt+=tran(k);}if(cnt<0) flag_ = false;}if(flag_&&cnt==0) flag = true;}}}if(flag) printf("YES\n");else printf("NO\n");}return 0;
}

Berland Crossword
题意给你一个n*n的矩阵 U,R,D,L分别代表上右下左四行中黑棋子的个数问你是否有一种方式可以构造出黑棋子符合在图中情况我们不难发现 n-2 是每一行最理想的情况，不会放在行的两边从而影响其他行，所以我们可以以这个为边界作为dfs跳出条件后用二进制枚举枚举四个边边角角放哪一个递归进行判断即可答案为所有递归答案取或

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl#define ll long long
#define fi first
#define se second
#define pr pair<int,int>const int MAX_N = 1000025;
bool check(int n,int U,int R,int D,int L,int sta)
{if(U<0||R<0||D<0||L<0) return false;int cnt = 0;cnt+=int(U>n-2)+int(R>n-2)+int(D>n-2)+int(L>n-2);if(cnt==0){return true;}else{bool flag = false;if((sta&1)==0){flag |= check(n,U-1,R,D,L-1,sta|1);}if((sta&2)==0){flag |= check(n,U-1,R-1,D,L,sta|2);}if((sta&4)==0){flag |= check(n,U,R-1,D-1,L,sta|4);}if((sta&8)==0){flag |= check(n,U,R,D-1,L-1,sta|8);}return flag;}}int main()
{int t,n,U,R,D,L;scanf("%d",&t);while(t--){scanf("%d%d%d%d%d",&n,&U,&R,&D,&L);if(check(n,U,R,D,L,0)) printf("YES\n");else printf("NO\n");}return 0;
}

1D Sokoban
题意给你n个数的数组与m个数的数组，每个数组中的数为坐标 -1e9<= x <= 1e9 且 x != 0 你可以在坐标0开始向左或者向右推被推的箱子会顺位到下一个位置问你操作完以后n和m个数的数组中值相等的情况最多有多少
这里有个trick：我们发现一种答案最大状态的一定是在你推到等于A中的箱子下标等于B中箱子的下标中 那么我们就进行枚举这个状态分别枚举推左边的时候的最大值你已经知道一连串箱子的长度（因为推了几个箱子会把他们连在一起成连续长块）再枚举右边的时候的最大值求最大值则是枚举左或右连续长块的左右下标二分得到包含了多少个B数组的数则为贡献注意统计不需要推原先已经匹配的数对这个过程的贡献用sum数组表示

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl#define ll long long
#define fi first
#define se second
#define pr pair<int,int>const int MAX_N = 200025;
int arr[MAX_N],brr[MAX_N],sum[MAX_N];
int main()
{int t;scanf("%d",&t);while(t--){int n,m;scanf("%d%d",&n,&m);arr[0] = -1;arr[n+1] = 2e9+1;for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),arr[i]+=1e9;brr[0] = -1;brr[m+1] = 2e9+1;for(int i = 1;i<=m;++i) scanf("%d",&brr[i]),brr[i]+=1e9;int xb = lower_bound(arr+1,arr+1+n,1e9)-arr;int r = xb,l = xb-1;int xb_ = lower_bound(brr,brr+1+m+1,1e9)-brr;int maxx_r = 0;sum[0] = 0;int k = 0;for(int i = 1;i<=m;++i){while(arr[k]<brr[i]) k++;if(brr[i]==arr[k]) sum[i] = sum[i-1] + 1;else sum[i] = sum[i-1];}for(int i = xb_;i<=m;++i){while(r+1<=n+1&&arr[r]<=brr[i]) r++;int L = brr[i]-(r-xb)+1;int LL = lower_bound(brr,brr+1+m+1,L)-brr-1;maxx_r = max(maxx_r,i-LL+sum[m]-sum[i]);}int maxx_l = 0;for(int i = xb_-1;i>0;--i){while(l-1>=0&&arr[l]>=brr[i]) l--;int R = brr[i]+(xb-1-l)-1;int RR = upper_bound(brr,brr+1+m+1,R)-brr-1;maxx_l = max(maxx_l,RR-i+1+sum[i-1]-sum[0]);}printf("%d\n",maxx_l+maxx_r);}return 0;
}

Dogeforces
题意老板和员工，老板至少有两个员工，员工不会有下级员工，但是老板可能有自己的老板，那么公司的创始人自然就是树上的顶点为最大的老板
nn的矩阵每个值 Ai,j 代表 i 员工与 j 员工 LCA(他们共同老板)的薪水，那么显然 Ai,i 代表第i个员工的薪水给你薪水矩阵构造出公司组织结构
Trick：我们发现创始人也就是最大的老板他是可以连无数条边到自己的下属于是我们将其单独抽出来为第 n+1 个公司的人也为公司创始人这个矩阵是角平分线对称的所以我们只需要考虑右上的情况那么最大有 500500/2(角平分线/2)个情况按照薪水少的关系去维护总共有如下4个情况：
用并查集维护这图并查集的fa数组代表集合中的老板ID
A B自己为员工，那么我们需要将这俩人连到一个老板，特判如果当前薪水为创始人薪水，则连到创始人
A B都不为员工，我们要将这俩人连到一个老板，但是如果这俩人其中一人是对方老板特判连接即可,特判如果当前薪水为创始人薪水，则连到创始人(但是可能其中一个人是创始人还得特判)
A B其中一位为员工，我们判断一下A,B中的老板薪水是不是当前枚举到的薪水，如果是则将员工连向另一个老板否则将他们连到一个新老板工资为创始人工资时连接到创始人

新老板我们用一个变量k表示 k初始值为n+2(n+1为创始人)

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl#define ll long long
#define fi first
#define se second
#define pr pair<int,int>const int MAX_N = 505;
const int MAX_M = MAX_N*MAX_N/2;
int arr[MAX_N][MAX_N];
int ans[MAX_M],fa[MAX_M];
priority_queue<pair<int,pair<int,int> > > q;
vector<pair<int,int> > vt;int Find(int x)
{if(fa[x]==x) return x;return fa[x] = Find(fa[x]);
}void Merge(int x,int y)
{x = Find(x), y = Find(y);if(x!=y) fa[x] = y;
}int main()
{int n,k,maxx = 0;scanf("%d",&n);for(int i = 1;i<MAX_M;++i) fa[i] = i;for(int i = 1;i<=n;++i){for(int j = 1;j<=n;++j){scanf("%d",&arr[i][j]);maxx = max(maxx,arr[i][j]);if(i==j) ans[i] = arr[i][j];else if(j>i){q.push(make_pair(-arr[i][j],make_pair(i,j)));}}}ans[n+1] = maxx;k = n+2;while(!q.empty()){pair<int,pair<int,int> > top = q.top();q.pop();int x = top.second.first, y = top.second.second;if(Find(x)==x&&Find(y)==y){if(-top.first==maxx){vt.push_back(make_pair(x,n+1));vt.push_back(make_pair(y,n+1));Merge(x,n+1);Merge(y,n+1);}else{vt.push_back(make_pair(x,k));vt.push_back(make_pair(y,k));ans[k] = -top.first;Merge(x,k);Merge(y,k);k++;}}else if(Find(x)!=x&&Find(y)!=y){int fa_x = Find(x),fa_y = Find(y);if(fa_x==fa_y) continue;if(maxx==-top.first){if(fa_x!=n+1) vt.push_back(make_pair(fa_x,n+1));if(fa_y!=n+1) vt.push_back(make_pair(fa_y,n+1));if(fa_x!=n+1) Merge(fa_x,n+1);if(fa_y!=n+1) Merge(fa_y,n+1);}else{if(ans[fa_x]==-top.first){vt.push_back(make_pair(fa_y,fa_x));Merge(fa_y,fa_x);}else if(ans[fa_y]==-top.first){vt.push_back(make_pair(fa_x,fa_y));Merge(fa_x,fa_y);}else{vt.push_back(make_pair(fa_x,k));vt.push_back(make_pair(fa_y,k));ans[k] = -top.first;Merge(fa_x,k);Merge(fa_y,k);k++;}}}else if(Find(x)!=x){int fa_x = Find(x);if(ans[fa_x] == -top.first){vt.push_back(make_pair(y,fa_x));Merge(y,fa_x);}else{if(-top.first==maxx){vt.push_back(make_pair(fa_x,n+1));vt.push_back(make_pair(y,n+1));Merge(fa_x,n+1);Merge(y,n+1);}else{vt.push_back(make_pair(fa_x,k));vt.push_back(make_pair(y,k));ans[k] = -top.first;Merge(fa_x,k);Merge(y,k);k++;}}}else if(Find(y)!=y){int fa_y = Find(y);if(ans[fa_y] == -top.first){vt.push_back(make_pair(x,fa_y));Merge(x,fa_y);}else{if(-top.first==maxx){vt.push_back(make_pair(fa_y,n+1));vt.push_back(make_pair(x,n+1));Merge(fa_y,n+1);Merge(x,n+1);}else{vt.push_back(make_pair(fa_y,k));vt.push_back(make_pair(x,k));ans[k] = -top.first;Merge(fa_y,k);Merge(x,k);k++;}}}}k--;printf("%d\n",k);for(int i = 1;i<=k;++i) i==k?printf("%d\n",ans[i]):printf("%d ",ans[i]);printf("%d\n",n+1);for(int i = 0;i<vt.size();++i){printf("%d %d\n",vt[i].first,vt[i].second);}return 0;
}

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