CF Round #580(div2)题解报告

T1 T2

水题,不管

T3

构造题,证明大约感性理解一下

我们想既然存在解

\(|a[n + i] - a[i]| = 1\)

这是必须要满足的

既然这样,那么图必须是这样的

\(-\),是相邻的两个数中的较小的一个,\(+\)是相邻的两个数中较大的

这样分配是肯定有解的

但是当n时偶数的时候,手玩一下就会发现,不可能满足+-交替,所以无解

#include<cstdio>

#include<iostream>

#include<queue>

#include<algorithm>

#include<cstring>

#include<cctype>

#include<vector>

#include<ctime>

#define LL long long

#define pii pair<int,int>

#define mk make_pair

#define fi first

#define se second

using namespace std;

inline int read(){

	int v = 0,c = 1;char ch = getchar();

	while(!isdigit(ch)){

		if(ch == '-') c = -1;

		ch = getchar();

	}

	while(isdigit(ch)){

		v = v * 10 + ch - 48;

		ch = getchar();

	}

	return v * c;

}

const int N = 5e5 + 3;

int a[N];

int n;

int main(){

	n = read();

	if(n & 1){

		int now = 0;

		for(int i = 1;i <= n;++i){

			if(!now) a[i] = i * 2 - 1,a[i + n] = i * 2;

			else a[i] = i * 2,a[i + n] = i * 2 - 1;

			now ^= 1;

		}

		printf("YES\n");

		for(int i = 1;i <= 2 * n;++i) printf("%d ",a[i]);

	}

	else printf("NO\n");

	return 0;

}

T4

分位考虑

这道题的关键在于给你张图,求最小环

由于点数和边数特别少,所以直接爆搜就好了

但是考虑多项式做法

最小环的本质是对于边\((u,v)\),去掉边后\((u,v)\)的最短路径\(+1\)

这可以在floyd的过程中搞一搞

当然,这是在边有边权的前提下

这道题目直接枚举每一条边然后bfs就好了

#include<cstdio>

#include<iostream>

#include<queue>

#include<algorithm>

#include<cstring>

#include<cctype>

#include<vector>

#include<ctime>

#define LL long long

#define pii pair<int,int>

#define mk make_pair

#define fi first

#define se second

const int INF = 0x3f3f3f3f;

using namespace std;

inline LL read(){

	LL v = 0,c = 1;char ch = getchar();

	while(!isdigit(ch)){

		if(ch == '-') c = -1;

		ch = getchar();

	}

	while(isdigit(ch)){

		v = v * 10 + ch - 48;

		ch = getchar();

	}

	return v * c;

}

const int N = 2e5 + 3;

LL a[N];

int n,ans;

int xx[65],yy[65];

int tot,rt;

struct edge{

	int to;

	int from;

	int nxt;

}e[N << 1];

int head[N];

int deep[N],fa[N];

int book[N];

int vis[N];

inline void add(int x,int y){

	e[++tot].to = y;

	e[tot].nxt = head[x];

	e[tot].from = x;

	head[x] = tot;

}

inline void dfs(int x,int dep){

//	book[x] = 1;

	vis[x] = 1;

	for(int i = head[x];i;i = e[i].nxt){

		int y = e[i].to;

		if(y == rt && dep > 2) ans = min(ans,dep);

		else if(!vis[y]) dfs(y,dep + 1);

	}

	vis[x] = 0;

}

int main(){

	//freopen("A.in","r",stdin);

	//freopen("A.out","w",stdout);

	n = read();

	for(int i = 1;i <= n;++i) a[i] = read();

	for(int i = 0;i <= 62;++i){

		int sum = 0;

		for(int j = 1;j <= n;++j){

			if(a[j] & (1ll << i)){

				if(!xx[i]) xx[i] = j;

				else if(!yy[i]) yy[i] = j;

				sum++;

				if(sum == 3){

					printf("3\n");

					return 0;

				}

			}

		}

		if(sum == 2) add(xx[i],yy[i]),add(yy[i],xx[i]);//printf("%d %d\n",xx[i],yy[i]);

	}

	ans = INF;

	for(int i = 1;i <= n;++i){

		dfs(rt = i,1);

		if(ans == 3){

			printf("3\n");return 0;

		}

	}

	if(ans > n) printf("-1\n");

	else printf("%d\n",ans);

	return 0;

}

CF Round #580(div2)题解报告的相关教程结束。