Codeforces Round #668 (Div. 2)

传送门：https://codeforces.com/contest/1405

A

将序列翻转输出即可。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb push_back
#define eb emplace_back
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/(b)

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

int main(){
	int T; cin>>T;
	while(T--){
		int n; cin>>n;
		vi res;
		rep(i,1,n){
			int v; read(v);
			res.pb(v);
		}
		reverse(all(res));
		for(auto i: res) cout<<i<<' ';
		cout<<endl;
	}
	return 0;
}

B

维护当前的正值，看看最后的正值是多少。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb push_back
#define eb emplace_back
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/(b)

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(ll &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

int main(){
	int T; cin>>T;
	while(T--){
		int n; cin>>n;
		ll cur=0;
		rep(i,1,n){
			ll v; read(v);
			cur=max(0LL, cur+v);
		}
		cout<<cur<<endl;
	}
	return 0;
}

C

注意到必须有第 (i) 个元素等于 (i+k) 个元素，如果出现冲突输出 NO，否则看看 ([1, k]) 的元素是否出现 (1) 或者 (0) 个数超过 (frac{k}{2}) 的情况，没有就是 YES。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb push_back
#define eb emplace_back
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/(b)

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

string s;

int main(){
	ios::sync_with_stdio(false);
	int T; cin>>T;
	while(T--){
		int n, k; cin>>n>>k;
		cin>>s;
		s=' '+s;
		for(auto &i: s) if(i=='?') i='2';
		
		vi res(n+1, 2);
		bool ok=true;
		rep(i,1,k) for(int j=i; j<=n; j+=k){
			int v=s[j]-'0';
			if(v==2) continue;
			if((v^res[i])==1){
				ok=false;
				break;
			}
			res[i]=v;
		}
		
		if(!ok){
			puts("NO");
			continue;
		}
		
		int cnt[2]={0};
		rep(i,1,n) if(res[i]!=2) cnt[res[i]]++;
		if(cnt[0]>k/2 || cnt[1]>k/2) puts("NO");
		else puts("YES");
	}
	return 0;
}

D

图论 + 结论题
Alice 胜利当且仅当下面至少一条满足：

• 树的直径 (geq 2da)
• (2dageq db)
• 初始距离 (dageq dis(a, b))

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb push_back
#define eb emplace_back
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/(b)

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=1e5+5, M=N<<1;

struct Edge{
	int to, next;
}e[M];

int n;
int h[N], tot;

void add(int u, int v){
	e[tot].to=v, e[tot].next=h[u], h[u]=tot++;
}

int fa[N], son[N], sz[N], dep[N];
int cnt, id[N], top[N];

void dfs1(int u, int father, int depth){
    fa[u]=father, sz[u]=1, dep[u]=depth;
    for(int i=h[u]; ~i; i=e[i].next){
        int go=e[i].to;
        if(go==father) continue;
        dfs1(go, u, depth+1);
        sz[u]+=sz[go];
        if(sz[go]>sz[son[u]]) son[u]=go;
    }
}

void dfs2(int u, int t){
    id[u]=++cnt, top[u]=t;
    if(!son[u]) return;
    dfs2(son[u], t);
    for(int i=h[u]; ~i; i=e[i].next){
        int go=e[i].to;
        if(go==fa[u] || go==son[u]) continue;
        dfs2(go, go);
    }
}

int lca(int u, int v){
    while(top[u]!=top[v]){
        if(dep[top[u]]<dep[top[v]]) swap(u, v);
        u=fa[top[u]];
    }
    return dep[u]<dep[v]? u: v;
}

int dis(int x, int y){
	return dep[x]+dep[y]-2*dep[lca(x, y)];
}

bool vis[N];
int d[N];

int find(int u){
	queue<int> q;
	q.push(u);
	rep(i,1,n) vis[i]=0, d[i]=INF;
	d[u]=0; int res=0;
	vis[u]=true;
	
	while(q.size()){
		int t=q.front(); q.pop();
		for(int i=h[t]; ~i; i=e[i].next){
			int go=e[i].to;
			if(!vis[go]){
				vis[go]=true;
				d[go]=d[t]+1;
				res=max(res, d[go]);
				q.push(go);
			}
		}
	}
	
	return res;
}

int get_D(){
	int u=1;
	rep(i,1,n) if(dep[i]>dep[u]) u=i;
	return find(u);
}

int main(){
	int T; cin>>T;
	while(T--){
		int a, b, x, y; cin>>n>>a>>b>>x>>y;
		rep(i,1,n) h[i]=-1, son[i]=fa[i]=top[i]=sz[i]=dep[i]=id[i]=0;
		tot=cnt=0;
		rep(i,1,n-1){
			int u, v; read(u), read(v);
			add(u, v), add(v, u);
		}
		
		dfs1(1, -1, 1), dfs2(1, 1);
		
		int d=get_D(); 
		if(dis(a, b)<=x || d<=2*x || 2*x>=y) puts("Alice");
		else puts("Bob");
	}
	return 0;
}