【AtCoder】ARC095 C-F题解

我居然每道题都能想出来

虽然不是每道题都能写对，debug了很久/facepalm

C - Many Medians

排序后前N/2个数的中位数时排序后第N/2 + 1的数

其余的中位数都是排序后第N / 2的数

#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <ctime>
#include <algorithm>
#include <map>
#define MAXN 200005
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ba 823
#define mo 974711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
int N;
int id[MAXN],X[MAXN],B[MAXN];
bool cmp(int a,int b) {
	return X[a] < X[b];
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    scanf("%d",&N);
    for(int i = 1 ; i <= N ; ++i) {
    	scanf("%d",&X[i]);
    	id[i] = i;
    }
    sort(id + 1,id + N + 1,cmp);
    int T = N / 2;
    for(int i = 1 ; i <= T ; ++i) {
    	B[id[i]] = X[id[T + 1]];
    }
    for(int i = T + 1 ; i <= N ; ++i) {
    	B[id[i]] = X[id[T]];
    }
    for(int i = 1 ; i <= N ; ++i) {
    	printf("%d\n",B[i]);
    }
    return 0;
}

D - Binomial Coefficients

N取最大的，R取最接近N/2的

#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <cstring>
#include <ctime>
#include <map>
#include <algorithm>
#include <cmath>
#define MAXN 200005
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ba 823
#define mo 974711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;

int N,A[100005];
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    ios::sync_with_stdio(false);
	scanf("%d",&N);
	for(int i = 1 ; i <= N ; ++i) scanf("%d",&A[i]);
	sort(A + 1,A + N + 1);
	printf("%d ",A[N]);
	double T = A[N] / 2.0;
	int t = 1;
	for(int i = 2 ; i <= N ; ++i) {
		if(fabs(A[i] - T) < fabs(A[t] - T)) t = i;
	}
	printf("%d\n",A[t]);
    return 0;
}

E - Symmetric Grid

事实上我想的优化不用也是个奇快无比的搜索

复杂度是啥不知道，反正1ms

发现两行能匹配上一定是两行的字符集是相同的

两列能匹配上一定是两列的字符集相同

我们枚举某行作为第一行，然后找它字符集相同的一行作为最后一行，枚举匹配，然后每一列都固定了，最后给剩下的N - 2行找匹配就行

匹配条件是某一行正着读等于另一行反着读

特判奇数行奇数列和H = 1，W = 1

#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <cstring>
#include <ctime>
#include <map>
#include <algorithm>
#include <cmath>
#define MAXN 200005
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ba 823
#define ha 99994711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
int H,W,St;
int Row[14][26],Col[14][26];
int matR[14][14],matC[14][14],idC[14],idR[14],mat[14];
bool vis[14],used[14];
char S[14][14];
void swapRow(int x,int y) {
	if(x == y) return;
	swap(idR[x],idR[y]);
	for(int i = 1 ; i <= W ; ++i) swap(S[x][i],S[y][i]);
}
void swapCol(int x,int y) {
	if(x == y) return;
	swap(idC[x],idC[y]);
	for(int i = 1 ; i <= H ; ++i) swap(S[i][x],S[i][y]);
}
bool match(int c) {
	if(c == 0) {
		memset(mat,0,sizeof(mat));
		int cnt = 0;
		for(int i = 2 ; i <= H - 1; ++i) {
			if(mat[i]) continue;
			for(int j = i + 1 ; j <= H - 1; ++j) {
				if(mat[j]) continue;
				bool f = 1;
				for(int k = 1 ; k <= W ; ++k) {
					if(S[i][k] != S[j][W - k + 1]) {
						f = 0;
						break;
					}
				}
				if(f) {mat[i] = j;mat[j] = i;break;}
			}
			if(!mat[i] && cnt < 1 && (H & 1)) {
				++cnt;
				for(int k = 1 ; k <= W ; ++k) {
					if(S[i][k] != S[i][W - k + 1]) return 0;
				}
				mat[i] = i;
			}
			else if(!mat[i]) return 0;
		}
		return 1;
	}
	if(c == W - c + 1) {
		for(int i = 1 ; i <= W ; ++i) {
			if(vis[i]) continue;
			if(S[1][i] == S[H][i]) {
				swapCol(i,c);
				vis[c] = 1;
				if(match(c - 1)) return 1;
				vis[c] = 0;
				swapCol(i,c);
			}
		}
	}
	else {
		for(int i = 1; i <= W; ++i) {
			if(vis[i]) continue;
			if(matC[idC[c]][idC[i]]) {
				if(S[1][c] == S[H][i] && S[1][i] == S[H][c]) {
					vis[c] = vis[W - c + 1] = 1;
					swapCol(i,W - c + 1);
					if(match(c - 1)) return 1;
					swapCol(i,W - c + 1);
					vis[c] = vis[W - c + 1] = 0;
				}
			}
		}
	}
	return 0;
}
bool check() {
	for(int i = 2 ; i <= H ; ++i) {
		if(matR[idR[1]][idR[i]]) {
			if(used[idR[i]]) continue;
			swapRow(i,H);
			if(match((W & 1) ? W / 2 + 1 : W / 2)) return 1;
			swapRow(i,H);
		}
	}
	return 0;
}
void Solve() {
	scanf("%d%d",&H,&W);
	for(int i = 1 ; i <= H ; ++i) scanf("%s",S[i] + 1);
	for(int i = 1 ; i <= H ; ++i) {
		for(int j = 1 ; j <= W ; ++j) {
			Row[i][S[i][j] - 'a']++;
		}
	}
	for(int j = 1 ; j <= W ; ++j) {
		for(int i = 1 ; i <= H ; ++i) {
			Col[j][S[i][j] - 'a']++;
		}
	}
	if(H == 1) {
		int cnt = 0;
		for(int i = 0 ; i <= 25 ; ++i) {
			if(Row[1][i] & 1) ++cnt;
		}
		if(cnt > 1) puts("NO");
		else puts("YES");
		return;
	}
	else if(W == 1) {
		int cnt = 0;
		for(int i = 0 ; i <= 25 ; ++i) {
			if(Col[1][i] & 1) ++cnt;
		}
		if(cnt > 1) puts("NO");
		else puts("YES");
		return;
	}
	for(int i = 1 ; i <= H ; ++i) {
		for(int j = i ; j <= H ; ++j) {
			bool f = 1;
			for(int k = 0 ; k <= 25 ; ++k) {
				if(Row[i][k] != Row[j][k]) {f = 0;break;}
			}
			matR[i][j] = matR[j][i] = f;

		}
	}
	for(int i = 1 ; i <= W ; ++i) {
		for(int j = i ; j <= W ; ++j) {
			bool f = 1;
			for(int k = 0 ; k <= 25 ; ++k) {
				if(Col[i][k] != Col[j][k]) {f = 0;break;}
			}
			matC[i][j] = matC[j][i] = f;
		}
	}
	for(int i = 1 ; i <= H ; ++i) idR[i] = i;
	for(int i = 1 ; i <= W ; ++i) idC[i] = i;
	for(int i = 1 ; i <= H ; ++i) {
		used[i] = 1;
		swapRow(i,1);
		if(check()) {puts("YES");return;}
		swapRow(i,1);
	}

	puts("NO");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - Permutation Tree

合法的只有一条链上长叶子

把叶子删掉，搜出一条链，链的两端如果有叶子就长出来

然后用一个数组记录链上每个点长的叶子的个数

然后从左右开始比较，如果左边小于右边就跳出，如果右边小于左边就翻转后跳出

构造的时候一个长了k个叶子的点排序就是 p + 1,p + 2...p + k, p

代码能力不行，跪了三次QAQ

#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <cstring>
#include <ctime>
#include <map>
#include <algorithm>
#include <cmath>
#define MAXN 100005
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ba 823
#define ha 99994711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
int N;
struct node {
	int to,next;
}E[MAXN * 2];
int sumE,head[MAXN],ind[MAXN],fa[MAXN],st,ed,D[MAXN];
int line[MAXN],tot;
void add(int u,int v) {
	E[++sumE].to = v;
	E[sumE].next = head[u];
	head[u] = sumE;
}
void dfs(int u) {
	ed = u;
	for(int i = head[u] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(v != fa[u] && ind[v] != -1) {
			fa[v] = u;
			dfs(v);
		}
	}
}
void Solve() {
	scanf("%d",&N);
	int u,v;
	for(int i = 1 ; i < N ; ++i) {
		scanf("%d%d",&u,&v);
		add(u,v);add(v,u);
		ind[u]++;ind[v]++;
	}
	if(N == 2) {
		puts("1 2");return;
	}
	memcpy(D,ind,sizeof(ind));
	for(int i = 1 ; i <= N ; ++i) {
		if(ind[i] == 1) {
			for(int j = head[i] ; j ; j = E[j].next) {
				int v = E[j].to;
				D[v]--;
			}
			D[i] = -1;
		}
	}
	memcpy(ind,D,sizeof(ind));
	for(int i = 1 ; i <= N ; ++i) {
		if(ind[i] > 2) {puts("-1");return;}
	}
	for(int i = 1 ; i <= N ; ++i) {
		if(ind[i] == 1) {st = i;dfs(st);break;}
	}
	if(!st) {
		for(int i = 1 ; i <= N ; ++i) {
			if(ind[i] == 0) {st = i;dfs(st);break;}
		}
	}
	for(int i = head[ed] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(v != fa[ed]) {
			fa[v] = ed;
			ed = v;
			break;
		}
	}

	for(int i = head[st] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(!fa[v]) {
			fa[st] = v;
			st = v;
			break;
		}
	}
	ind[ed] = ind[st] = 1;
	int p = ed;
	while(p) {
		++tot;
		for(int i = head[p] ; i ; i = E[i].next) {
			int v = E[i].to;
			if(ind[v] == -1) {
				line[tot]++;
			}
		}
		p = fa[p];
	}
	int L = 1,R = tot;
	while(L <= tot && R >= 1) {
		if(line[L] < line[R]) break;
		else if(line[L] > line[R]) {reverse(line + 1,line + tot + 1);break;}
		++L;--R;
	}
	p = 2;
	printf("%d",1);
	for(int i = 2 ; i <= tot ; ++i) {
		for(int j = p + 1 ; j <= p + line[i] ; ++j) {
			printf(" %d",j);
		}
		printf(" %d",p);
		p = p + line[i] + 1;
	}
	putchar('\n');
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}