Starry

发表于 更新于 分类于 本文字数： 627 阅读时长 ≈ 2 分钟

这题很容易想到的一点是对于格子中只有一个是，并且我们填一个字母有可能使得它变成 check patterns的话，我们要先把这种格子填什么确定下来，而且这是唯一的。

接下来就是处理它周围格子了，我们就必须按着遍历顺序相反的方向开始弄上面这类格子，相同方向的接下来会遍历到。

那么如果不存在上述格子，且仍还有的格子，那我们就先随便填一种，然后check重复上面的过程

最后的工作就是检查是否存在check patterns了

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#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

ll tt,n,m,a[(int)109][(int)109] ;
char c;

void dfs(int x,int y){
	if(x<=0 || y<=0 || x>=n || y>=m)return;
	ll p=-1,q=-1,cnt=0;
	for(int i=0;i<=1;++i)for(int j=0;j<=1;++j)cnt+=a[x+i][y+j]==1;
	if(cnt!=1)return;
	for(int i=0;i<=1;++i)for(int j=0;j<=1;++j)if(a[x+i][y+j]==1){p=i,q=j;}
	if(a[x+!p][y+q]==a[x+p][y+!q]&&(a[x+!p][y+q]^a[x+!p][y+!q])==1){
		a[x+p][y+q]=a[x+!p][y+q];
		for(int i=0;i<=1;++i)for(int j=0;j<=1;++j)dfs(x+p-i,y+q-j);//back
	}
}

bool solve(){
	for(int i=1;i<n;++i){
		for(int j=1;j<m;++j){
			dfs(i,j);
		}
	}
	for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){
		if(a[i][j]==1){
			a[i][j]=3;
			for(int k=0;k<=1;++k)for(int l=0;l<=1;++l)dfs(i-k,j-l);
		}
	}
	for(int i=1;i<n;++i)for(int j=1;j<m;++j){
		if(a[i][j]==a[i+1][j+1]&&a[i+1][j]==a[i][j+1]&&a[i][j]!=a[i+1][j])return false;
	}
	return true;
}

int main()
{
    for(scanf("%lld",&tt);tt;tt-=1){
    	scanf("%lld%lld",&n,&m);
    	for(int i=1;i<=n;++i){
    		for(int j=1;j<=m;++j){
    			scanf(" %c",&c);
    			if(c=='?')a[i][j]=1;
    			else if(c=='B')a[i][j]=2;
    			else a[i][j]=3;
    		}
    		a[i][m+1]=0;
    	}
    	for(int i=1;i<=m+1;++i)a[n+1][m]=0;
    	if(solve()){
    		printf("%s","YES\n");
    		for(int i=1;i<=n;++i){
    			for(int j=1;j<=m;++j)printf("%c",a[i][j]==2?'B':'W');
    			printf("\n");
    		}
    	}else printf("%s","NO\n");
    }
    return 0;
}

发表于 更新于 分类于 本文字数： 254 阅读时长 ≈ 1 分钟

考虑到树状数组是一颗树，然后染色的做法

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#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

int tt,n,a[(int)1e5+9],ans,cn[(int)1e5+9];
char c;
vi g[(int)1e5+9]; 

void dfs(int x){
	a[x]=0;
	bool ok=1;
	for(auto it:g[x]){
		if(a[it]){
			dfs(it);
			ok=0;
		}
	}
	ans+=ok;
}

int main()
{
    for(int i=1;i<=1e5;++i){
    	int tmp=i+lowbit(i);
    	if(tmp<=1e5)g[tmp].push_back(i);
    }
    scanf("%d",&tt);
    f(sb,1,tt){
    	ans=0;
    	scanf("%d",&n);
    	for(int i=1;i<=n;++i)scanf(" %c",&c),a[i]=c=='1',cn[i]=0;
    	for(int i=n;i;--i){
    		if(a[i]){
    			dfs(i);
    			int tmp=i+lowbit(i);
    			if(tmp<=n)cn[tmp]+=1;
    		}
    	}
    	for(int i=1;i<=n;++i)ans+=cn[i]==1;
    	printf("%d\n",ans);
    }
    return 0;
}

发表于 更新于 分类于 本文字数： 460 阅读时长 ≈ 2 分钟

分情况讨论：如果的话，就直接跑Dijkstra。否则就先用优先堆模拟出可以到达哪些点，然后对这些点跑Dijkstra，不过dp方程要改变

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#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x & (-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;

using ll = long long;

ll tt, n, m, A, B, l[(int)2e5 + 9], dis[(int)2e5 + 9];
vi g[(int)2e5 + 9];
bitset<(int)2e5 + 9> vis;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> tt;
	f(sb, 1, tt)
	{
		cin >> n >> m >> A >> B;
		for (int i = 1; i <= n; ++i)
			g[i].clear(), dis[i] = 0x3f3f3f3f;
		for (int i = 1, x, y; i <= m; ++i)
		{
			cin >> x >> y;
			g[x].push_back(y), g[y].push_back(x);
		}
		for (int i = 1; i <= n; ++i)
			cin >> l[i], l[i] += i == 1 ? 0ll : B;
		if (A <= B)
		{
			queue<int> qu;
			qu.push(1);
			dis[1] = 0;
			while (!qu.empty())
			{
				auto p = qu.front();
				qu.pop();
				// cerr<<p<<"::"<<dis[p]<<"\n";
				for (auto it : g[p])
				{
					// cerr<<it<<":::"<<((dis[p]+1)*(A-B)+l[1])<<"\n";
					if (dis[it] > dis[p] + 1 && dis[p] * (A - B) + l[1] > l[it])
					{
						dis[it] = dis[p] + 1;
						qu.push(it);
					}
				}
			}
			dis[n] = dis[n] == 0x3f3f3f3f ? -1ll : dis[n];
			cout << dis[n] << "\n";
		}
		else
		{
			priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
			vis.reset();
			ll day = -1;
			q.push({0,1});
			while (!q.empty())
			{
				auto [aa, bb] = q.top();
				q.pop();
				if (vis[bb])
					continue;
				if(day!=-1&&aa>=l[1]+day*(A-B))break;
				vis[bb]=1,day+=1;
				for (auto it : g[bb])
				{
					if (!vis[it])
						q.push({l[it], it});
				}
			}
			q.push({0, 1});
			dis[1] = 0;
			while (!q.empty())
			{
				auto [aa, bb] = q.top();
				q.pop();
				if (aa>dis[bb])
					continue;
				for (auto it : g[bb])
				{
					if(vis[it]&&dis[it]>aa+1&&l[it]<l[1]){
						dis[it]=aa+1;
						q.push({dis[it],it});
					}else if(vis[it]&&dis[it]>max(aa+1ll,(l[it]-l[1])/(A-B)+2)){
						dis[it]=max(aa+1ll,(l[it]-l[1])/(A-B)+2);
						q.push({dis[it],it});
					}
				}
			}
			dis[n]=dis[n]==0x3f3f3f3f?-1ll:dis[n];
			cout << dis[n] << "\n";
		}
	}
}

发表于 更新于 分类于 本文字数： 195 阅读时长 ≈ 1 分钟

amazing

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#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

ll tt,n,a[(int)1e5+9],b[(int)1e5+9]; 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>tt;
    f(sb,1,tt){
    	cin>>n;
    	f(i,1,n)cin>>a[i];
    	f(i,1,n)cin>>b[i];
    	auto check=[&](ll val){
    		vector<ll> x,y;
    		f(i,1,n)x.push_back(a[i]&val),y.push_back(~b[i]&val);
    		sort(all(x)),sort(all(y));
    		return x==y;
    	};
    	ll ans=0;
    	for(int i=29;~i;--i){
    		if(check(ans|(1ll<<i))){
    			ans|=(1ll<<i);
    		}
    	}
    	cout<<ans<<"\n";
    }
    return 0;
}

发表于 更新于 分类于 本文字数： 262 阅读时长 ≈ 1 分钟 
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#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

int n,q;
bitset<40> a[(int)1e5+9],b[(int)1e5+9];
vector<pii> v[(int)1e5+9];

void zero(int x,int y,int val){
	for(int i=0;i<30;++i){
		if(!((1ll<<i)&val)){
			a[x][i]=a[y][i]=1;
		}
	}
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>q;
    for(int i=1,x,y,z;i<=q;++i){
    	cin>>x>>y>>z;
    	if(x==y){
    		b[x]=z;
    		zero(x,x,z);
    		continue;
    	}
    	v[x].push_back({y,z});
    	v[y].push_back({x,z});
    	zero(x,y,z);
    }
    for(int i=1;i<=n;++i){
    	for(auto [aa,bb]:v[i]){
    		bitset<40> tt(bb);
    		b[i]|=(a[aa]&tt);
    	}
    }
    for(int i=1;i<=n;++i){
    	for(auto [aa,bb]:v[i]){
    		if(i>aa){
    			bitset<40> tt(bb);
    			b[i]|=(tt^b[aa]);
    		}
    	}
    }
    for(int i=1;i<=n;++i){
    	cout<<b[i].to_ullong()<<" \n"[i==n];
    }
    return 0;
}
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