AtCoder Beginner Contest 302

A - Attack

Problem Statement

题意：敌人有$A$的耐力值，每次攻击敌人可以减少$B$的耐力值，问多少次敌人耐力值$<=0$?

Solution

题解:$\dfrac{a}{b}$上取整。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
	ll a,b;
	cin>>a>>b;
	if(a%b==0)
		cout<<a/b<<endl;
	else cout<<a/b+1<<endl;
	return 0;
}

B - Find snuke

Problem Statement

题意：给你一个$n$行$m$列矩阵，里面都是小写英文字母，输出在同一直线上出现的$snuke$。输出坐标。

Solution

题解:遍历一遍，找到$s$的位置进行$check$。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 110;
char a[N][N];
int n,m;
bool check(int i,int j)
{
	bool ok = false;
	if(i-4>=1)
		if(a[i-1][j]=='n'&&a[i-2][j]=='u'&&a[i-3][j]=='k'&&a[i-4][j]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i-1<<" "<<j<<endl;
			cout<<i-2<<" "<<j<<endl;
			cout<<i-3<<" "<<j<<endl;
			cout<<i-4<<" "<<j<<endl;
			return ok;
		}
	if(i+4<=n)
		if(a[i+1][j]=='n'&&a[i+2][j]=='u'&&a[i+3][j]=='k'&&a[i+4][j]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i+1<<" "<<j<<endl;
			cout<<i+2<<" "<<j<<endl;
			cout<<i+3<<" "<<j<<endl;
			cout<<i+4<<" "<<j<<endl;
			return ok;
		}
	if(j-4>=1)
		if(a[i][j-1]=='n'&&a[i][j-2]=='u'&&a[i][j-3]=='k'&&a[i][j-4]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i<<" "<<j-1<<endl;
			cout<<i<<" "<<j-2<<endl;
			cout<<i<<" "<<j-3<<endl;
			cout<<i<<" "<<j-4<<endl;
			return ok;
		}
	if(j+4<=m)
		if(a[i][j+1]=='n'&&a[i][j+2]=='u'&&a[i][j+3]=='k'&&a[i][j+4]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i<<" "<<j+1<<endl;
			cout<<i<<" "<<j+2<<endl;
			cout<<i<<" "<<j+3<<endl;
			cout<<i<<" "<<j+4<<endl;
			return ok;
		}
	if(i-4>=1&&j-4>=1)
		if(a[i-1][j-1]=='n'&&a[i-2][j-2]=='u'&&a[i-3][j-3]=='k'&&a[i-4][j-4]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i-1<<" "<<j-1<<endl;
			cout<<i-2<<" "<<j-2<<endl;
			cout<<i-3<<" "<<j-3<<endl;
			cout<<i-4<<" "<<j-4<<endl;
			return ok;
		}
	if(i+4<=n&&j+4<=m)
		if(a[i+1][j+1]=='n'&&a[i+2][j+2]=='u'&&a[i+3][j+3]=='k'&&a[i+4][j+4]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i+1<<" "<<j+1<<endl;
			cout<<i+2<<" "<<j+2<<endl;
			cout<<i+3<<" "<<j+3<<endl;
			cout<<i+4<<" "<<j+4<<endl;
			return ok;
		}
	if(i-4>=1&&j+4<=m)
		if(a[i-1][j+1]=='n'&&a[i-2][j+2]=='u'&&a[i-3][j+3]=='k'&&a[i-4][j+4]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i-1<<" "<<j+1<<endl;
			cout<<i-2<<" "<<j+2<<endl;
			cout<<i-3<<" "<<j+3<<endl;
			cout<<i-4<<" "<<j+4<<endl;
			return ok;
		}
	if(i+4<=n&&j-4>=1)
		if(a[i+1][j-1]=='n'&&a[i+2][j-2]=='u'&&a[i+3][j-3]=='k'&&a[i+4][j-4]=='e')
		{
			ok = true;
			cout<<i<<" "<<j<<endl;
			cout<<i+1<<" "<<j-1<<endl;
			cout<<i+2<<" "<<j-2<<endl;
			cout<<i+3<<" "<<j-3<<endl;
			cout<<i+4<<" "<<j-4<<endl;
			return ok;
		}

	return ok;

}
int main()
{
	cin>>n>>m;
	for(int i = 1;i<=n;i++)
		for(int j = 1;j<=m;j++)
			cin>>a[i][j];
	for(int i = 1;i<=n;i++)
	{
		for(int j = 1;j<=m;j++)
		{
			if(a[i][j]=='s')
			{
				if(check(i,j))return 0;
			}
		}
	}
	return 0;
}

C - Almost Equal

Problem Statement

题意：给你$N$个长度为$M$的字符串，问你能否对这些字符串重新排列，使得第$i$个和第$i+1$个字符串只有一个字母不一样，可以输出$Yes$,否则$No$.

Solution

题解:因为注意到$N$范围很小，可以直接暴力$dfs$，先进性一个全排列，每次$check$是否满足。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,cnt;
const int N = 10;
string s[N];
int a[N];
bool vis[N];
map<int,string>mp;
bool yes;
void dfs(int step)
{
	if(yes)return;
	if(step>n)
	{
		bool ok = true;
		for(int i = 1;i<n;i++)
		{
			int c = 0;
			string t = mp[a[i]];
			string t2 = mp[a[i+1]];
			for(int j = 0;j<m;j++)
			{
				if(t[j]!=t2[j])c++;
			}
			if(c!=1)
			{
				ok = false;
				break;
			}
		}
		if(ok)yes = true;
		return;
	}	
	for(int i = 1;i<=n;i++)
	{
		if(!vis[i])
		{
			a[step] = i;
			vis[i] = true;
			dfs(step+1);
			vis[i] = false;
			a[step] = 0;
		}
	}
}
int main()
{
	cin>>n>>m;
	for(int i = 1;i<=n;i++)
		cin>>s[i],mp[++cnt] = s[i];
	dfs(1);
	if(yes)cout<<"Yes\n";
	else cout<<"No\n";
	return 0;
}

D - Impartial Gift

Problem Statement

题意：要为两个人选礼物，从$N$个候选中选一个给$A$，从$M$个候选中选一个给$B$,问两个的价值相差不超过$D$，问你两个礼物总和最大为多少？

Solution

题解:排序+双指针。因为我们优先选价值高的，如果不满足价值差小于$D$就移动价值大的那个指针，因为这个时候移动小的只会更厘普。

注意$A,B$的范围，小心爆$long long$

#include<bits/stdc++.h>
using namespace std;
__int128 read()
{
	__int128 f = 1, w = 0;
	char ch = getchar();
	while (ch < '0' || ch>'9')
	{
		if (ch == '-')
			f = -1;
		ch = getchar();
	}
	while (ch <= '9' && ch >= '0')
	{
		w = w * 10 + ch - '0';
		ch = getchar();
	}
	return f * w;
}

void print(__int128 x)
{
	if (x < 0)
	{
		putchar('-');
		x = -x;
	}
	if (x > 9)print(x / 10);
	putchar(x % 10 + '0');
}

typedef __int128 ll;
const int N = 2e5+10;
ll n,m,D;
ll a[N],b[N];
bool cmp(ll a,ll b)
{
	return a>b;
}
int main()
{
	n = read(),m = read(),D = read();
	for(int i = 1;i<=n;i++)
		a[i] = read();
	for(int i = 1;i<=m;i++)
		b[i] = read();
	sort(a+1,a+1+n,cmp);
	sort(b+1,b+1+m,cmp);
	ll c = 1,d = 1;
	bool ok = false;
	ll ans = 0;
	while(c<=n&&d<=m)
	{

		if((a[c]-b[d]>=0&&a[c]-b[d]<=D)||(b[d]-a[c]>=0&&b[d]-a[c]<=D))
		{
			ok = true;
			ans = a[c]+b[d];
			break;
		}
		else
		{
			if(a[c]>b[d])c++;
			else d++;
		}
	}
	if(ok)
		print(ans);
	else cout<<-1<<endl;
	return 0;
}

E - Isolation

Problem Statement

题意:给你一个无向图，$N$个点，编号$1$到$N$,并且初始状态没有边。

给你$Q$个询问，每次询问输出当前孤立的点的数目。

$1$ $ u$ $ v$ :连接$uv$

$2 $ $ v$ ：删除和$v$相连的所有边

Solution

题解：用$set$存边，先说操作一：对于点x,y,连接他们两个，如果set[x].size()==0的话，那现在连上了，那答案-1，同样对于$y$也是。对于操作二：删边操作。我们遍历该点连的所有边，用lower_bound直接erase掉对应点和$x$的边，如果删完之后edge[y].size()变为0了，那答案+1，最后如果对于操作的点$x$，如果edge[x].size是大于0的，那答案-1，edge[x].clear。解决。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e5+10;
int cnt;
map<pair<int,int>,int>mp;
set<int>edge[N];
int main()
{
	int n,q;
	cin>>n>>q;
	cnt = n;
	for(int i = 1;i<=q;i++)
	{
		int op;
		cin>>op;
		if(op==1)
		{
			int x,y;
			cin>>x>>y;	
			cnt -= edge[x].empty(),cnt-=edge[y].empty();
			edge[x].insert(y);
			edge[y].insert(x);
			cout<<cnt<<endl;
		}
		else
		{
			int x;
			cin>>x;
			for(auto y:edge[x])
			{
				edge[y].erase(edge[y].lower_bound(x));
				cnt += edge[y].empty();
			}
			if(edge[x].size())
				cnt++,edge[x].clear();
			cout<<cnt<<endl;
		}
	}
	return 0;
}

F - Merge Set

Problem Statement

题意:给你$N$个集合，我们每次选2个集合进行合并，这两个集合要至少有一个一样的元素才能合并，问至少合并多少次能使得集合中同时包含$1$和$M$。

Solution

思路：是01bfs。我们考虑把$N$个集合看成$N$个点，$M$个元素看成$M$个点，建图跑最短路。当前集合到集合内元素的距离设为0，集合元素到集合的距离设为1，跑bfs或者dijkstra都可以。因为是求合并次数，最后答案-1.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5+10;
struct Node
{
	int y,v;
	Node(int _y,int _v){y = _y;v = _v;};
};
set<pair<int,int>>q;
vector<Node>edge[N*2];
int n,m,k, dist[N*2];
inline void dijkstra(int s)
{
	memset(dist,127,sizeof(dist));
	dist[s] = 0;
	q.clear();
	for(int i = 1;i<=n+m;i++)
	{
		q.insert({dist[i],i});
	}
	while(!q.empty())
	{
		int x = q.begin()->second;//离起点最近的点的下标
		q.erase(q.begin());
		for(auto i:edge[x])
		{
			if(dist[x]+i.v<dist[i.y])
			{
				q.erase({dist[i.y],i.y});
				dist[i.y]= dist[x]+i.v;
				q.insert({dist[i.y],i.y});
			}
			
		}
	}
}


int main()
{
	cin>>n>>m;
	for(int i = 1;i<=n;i++)
	{
		int sz;
		cin>>sz;
		for(int j = 1;j<=sz;j++)
		{
			int x;
			cin>>x;
			edge[i].push_back(Node(x+n,0));
			edge[x+n].push_back(Node(i,1));
		}
	}
	dijkstra(1+n);
	ll ans = dist[m+n];
	if(ans>(1<<30))ans= 0;
	cout<<ans-1<<endl;
	return 0;
}

posted on 2023-06-17 08:54 nannandbk 阅读(40) 评论(0) 收藏举报

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