CSP-S2023题解

lock

直接模拟题意，过程略。

#include<bits/stdc++.h>
using namespace std;
int st[15][15];
int dis(int x,int y){
	if(x < y)return y - x;
	return y + 10 - x;
}
bool match(int a[],int b[]){
	int num = 0,s = 0,p = 0;
	for(int i = 1;i <= 5;i++)
		if(a[i] != b[i]){
			if(num == 0){
				num++;
				s = dis(a[i],b[i]);
				p = i;
			}
			else if(num == 1){
				if(i != p + 1)return false;
				if(dis(a[i],b[i]) != s)return false;
				num++;	
			}
			else return false;
		}
	return (num == 1 || num == 2);
}
int main(){
	int n;
	cin >> n;
	for(int i = 1;i <= n;i++)
		for(int j = 1;j <= 5;j++)cin >> st[i][j];
	int ans = 0;
	for(int i = 0;i <= 99999;i++){
		int t = i;int cur[15];
		for(int j = 1;j <= 5;j++){cur[j] = t % 10;t /= 10;}
		int mat = 0;
		for(int j = 1;j <= n;j++)
			if(match(cur,st[j]))mat++;
		if(mat == n)ans++;
	}
	cout << ans << '\n';
	return 0;
}

game

考虑平方怎么做。拿一个栈来匹配，每次栈顶能消就消，记录一下每个元素匹配的位置，令其为 \(to_i\)，那么一个元素如果没匹配上一定是跳过匹配的一段去匹配，也就是不断地令 \(j = to_j +1\) 直到匹配，算一下深度就是答案。

时间复杂度 \(O(26n)\)，原因是每种字符只会遍历一遍

#include<bits/stdc++.h>
using namespace std;
const int MX = 2e6 + 7;
int n;char s[MX];
int to[MX],dep[MX];
int main(){
	cin >> n >> s + 1;
	to[n] = n + 1;
	long long ans = 0;
	for(int i = n - 1;i >= 1;i--){
		int j = i + 1;
		while(j <= n && s[j] != s[i])j = to[j] + 1;
		to[i] = j;
		if(to[i] > n)continue;
		else{
			dep[i] = dep[to[i] + 1] + 1;
			ans += dep[i];
		}
	}
	cout << ans << '\n';
	return 0;
}

struct

模拟。代码逻辑较为清晰，建议直接看。

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define pll pair<long long,long long >
#define fir first
#define sec second
const int MX = 1005;

struct Type{
	string name;
	int shift;
	ll sz;
	vector<string > itemType,itemName;
	vector<ll > itemLpos,itemRpos;
	void print(){
		cerr << "struct name = " << name << '\n';
		cerr << "shift = " << shift << '\n';
		cerr << "sz = " << sz << '\n';
		for(int i = 0;i < itemType.size();i++){
			cerr << itemType[i] << ' ' << itemName[i] << '\n';
			cerr << "memory pos : " << itemLpos[i] << ' ' << itemRpos[i] << '\n';
		}
	}
}st[MX];
int id = 0;

int tot = 3;
map<string,int > nameTable;

void init(){
	nameTable["byte"] = 0;
	nameTable["short"] = 1;
	nameTable["int"] = 2;
	nameTable["long"] = 3;
	st[0].name = "byte";
	st[1].name = "short";
	st[2].name = "int";
	st[3].name = "long";
	st[0].shift = 1,st[0].sz = 1;
	st[1].shift = 2,st[1].sz = 2;
	st[2].shift = 4,st[2].sz = 4;
	st[3].shift = 8,st[3].sz = 8;
}

void opt1(){
	string name;int k,maxshift = 0;ll sz = 0;
	cin >> name >> k;
	vector<string > itTypes,itNames;
	for(int i = 1;i <= k;i++){
		string itType,itName;
		cin >> itType >> itName;
		itTypes.push_back(itType);
		itNames.push_back(itName);
		int id = nameTable[itType];
		maxshift = max(maxshift,st[id].shift);
	}
	int cur = ++tot;ll L = 0,lasSz = 0;
	for(int i = 0;i < k;i++){
		string itType = itTypes[i],itName = itNames[i];
		int id = nameTable[itType];
		if(i != 0){
			L += lasSz;
			while(L % st[id].shift != 0)L++;
		}
		st[cur].itemType.push_back(itType);
		st[cur].itemName.push_back(itName);
		st[cur].itemLpos.push_back(L);
		st[cur].itemRpos.push_back(L + st[id].sz - 1);
		lasSz = st[id].sz;
	}
	L += lasSz;
	while(L % maxshift != 0)L++;
	st[cur].shift = maxshift;st[cur].sz = L;
	nameTable[name] = cur;
	st[cur].name = name;
	cout << L << ' ' << maxshift << '\n';
}

struct mem{
	ll L,R;
	string type,name;
	bool operator < (const mem b)const{
		return L < b.L;
	}
};

mem memBox[MX];
int lasMem = 0,memcnt = 0;

void opt2(){
	string type,name;
	cin >> type >> name;
	int id = nameTable[type];
	if(lasMem != 0){
		while(lasMem % st[id].shift != 0)lasMem++;
	}
	cout << lasMem << '\n';
	mem newE;
	newE.L = lasMem;newE.R = lasMem + st[id].sz - 1;
	lasMem = newE.R + 1;
	newE.type = type;newE.name = name;
	memBox[++memcnt] = newE;
}

void opt3(string ele){
	vector<string > div;string tmp = "";
	for(int i = 0;i < ele.size();i++){
		if(ele[i] == '.'){
			div.push_back(tmp);
			tmp = "";
		}
		else tmp += ele[i];
	}
	div.push_back(tmp);
	tmp = "";
	int pos = 0,typeId = 0;
	for(int i = 1;i <= memcnt;i++){
		if(memBox[i].name == div[0]){
			pos = i;
			typeId = nameTable[memBox[i].type];
		}
	}
	//pos = ele pos
	ll dx = memBox[pos].L;
	for(int i = 1;i < div.size();i++){
		for(int j = 0;j < st[typeId].itemType.size();j++)
			if(st[typeId].itemName[j] == div[i]){
				dx += st[typeId].itemLpos[j];
				typeId = nameTable[st[typeId].itemType[j]];
				break;
			}
	}
	cout << dx << '\n';
}

void opt4(){
	ll addr,tmpd;
	cin >> addr;
	tmpd = addr;
	int typeId = 0;bool find = false;
	vector<string > output;
	for(int i = 1;i <= memcnt;i++)
		if(memBox[i].L <= addr && addr <= memBox[i].R){
			typeId = nameTable[memBox[i].type];
			output.push_back(memBox[i].name);
			addr -= memBox[i].L;
			find = true;
		}
	if(!find){
		cout << "ERR\n";
		return;
	}
	bool flag = true;
	for(;;){
		if(typeId < 4)break;
		bool ok = false;
		for(int j = 0;j < st[typeId].itemType.size();j++){
			if(st[typeId].itemLpos[j] <= addr && addr <= st[typeId].itemRpos[j]){
				addr -= st[typeId].itemLpos[j];
				output.push_back(st[typeId].itemName[j]);
				typeId = nameTable[st[typeId].itemType[j]];
				ok = true;
				break;
			}
		}
		if(!ok){
			flag = false;
			break;
		}
	}
	if(!flag)cout << "ERR\n";
	else{
		for(int i = 0;i < output.size();i++){
			if(i != 0)cout << ".";
			cout << output[i];
		}
		cout << '\n';
	}
	
}

signed main(){
	init();
	int n;
	cin >> n;
	for(int i = 1;i <= n;i++){
		int opt;
		cin >> opt;
		if(opt == 1)opt1();
		if(opt == 2)opt2();
		if(opt == 3){
			string tmp;
			cin >> tmp;
			opt3(tmp);
		}
		if(opt == 4)opt4();
	}
	//for(int i = 4;i <= tot;i++)st[i].print();
	return 0;
}

tree

考虑先二分答案，对于每个点求出至少要在什么时间踩上去，然后直接贪心，每次去标时间最靠前的节点，时间复杂度 \(O(n\log^2V)\) 至 \(O(n\log V)\)。

我写的 \(O(n\log n\log V)\)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define i128 __int128
i128 read(){
	i128 x = 0,ch = getchar(),f = 1;
	while(ch < 48 || ch > 57){
		if(ch == '-')f = -1;
		ch = getchar();
	}
	while(ch >= 48 && ch <= 57)x = x * 10 + ch - 48,ch = getchar();
	return x * f;
}

const int MX = 1e5 + 7;

int n;
ll a[MX],b[MX],c[MX],t[MX];
vector<int > g[MX];

//b,c,天数为d
i128 calc(ll b,ll c,ll d){
	if(c < 0){
		ll zero = (1 - b) / c;
		if((1 - b) % c == 0)zero--;
		i128 ret = 0,d0 = min(d,zero);
		ret += (i128)(b + c + b + (i128)d0 * c) * d0 / 2;
		ret += d - d0;
		return ret;
	}
	return (i128)(b + c + b + (i128)c * d) * d / 2;
}
//a,b,c的树最晚要啥时候踩上去
int days(ll a,ll b,ll c,int T){
	int l = 1,r = min((int)n,T),ans = 0;
	i128 tot = calc(b,c,T);
	while(l <= r){
		int mid = l + r >> 1;
		if(tot - calc(b,c,mid - 1) >= a){ans = mid,l = mid + 1;}
		else r = mid - 1;
	}
	return ans;
}

bool vis[MX];int fa[MX];

void dfs(int x,int f){
	fa[x] = f;
	for(int i = 0;i < g[x].size();i++){
		int to = g[x][i];
		if(to == f)continue;
		dfs(to,x);
	}
}

struct st{
	int id;
	i128 t;
}s[MX];

st mk(int id,i128 t){
	st ret;
	ret.id = id;ret.t = t;
	return ret;
}

vector<st > buc[MX];

bool cmp(st x,st y){return x.t < y.t;}

bool check(int T){
	for(int i = 1;i <= n;i++)t[i] = days(a[i],b[i],c[i],T);
	//cerr << "cur ending = " << (int)T << '\n';
	//cerr << "days : \n";
	//for(int i = 1;i <= n;i++)cout << (int)t[i] << ' ';
	//cout << '\n';
	for(int i = 1;i <= n;i++){buc[i].clear();buc[i].resize(0);}
	for(int i = 1;i <= n;i++)
		if(t[i] == 0)return false;
		else if(t[i] <= n)buc[t[i]].push_back(mk(i,t[i]));
	int tot = 0;
	for(int i = 0;i <= n;i++)
		for(int j = 0;j < buc[i].size();j++)s[++tot] = buc[i][j];
	memset(vis,false,sizeof(vis));
	int d = 1;
	//cerr << "start\n";
	for(int k = 1;k <= tot;k++){
		int i = s[k].id;
		if(vis[i])continue;
		vector<int > grand;
		grand.push_back(i);
		while(fa[i] != 0 && !vis[fa[i]]){
			grand.push_back(fa[i]);
			i = fa[i];
		}
		for(int j = grand.size() - 1;j >= 0;j--){
			//cerr << "col node " << grand[j] << '\n';
			if(t[grand[j]] < d)return false;
			vis[grand[j]] = true;
			d++;
		}
	}
	return true;
}

int main(){
	n = read();
	for(int i = 1;i <= n;i++){a[i] = read();b[i] = read();c[i] = read();}
	for(int i = 1;i < n;i++){
		int u = read();int v = read();
		g[u].push_back(v);g[v].push_back(u);
	}
	//cout << "node 4 : \n";
	//for(int i = 1;i <= 6;i++)cout << (int)calc(b[4],c[4],i) << ' ';
	//cerr << '\n';
	
	dfs(1,0);
	int l = 1,r = 1000000000,ans = 0;
	while(l <= r){
		int mid = l + r >> 1;
		if(check(mid)){ans = mid,r = mid - 1;}
		else l = mid + 1;
	}
	cout << ans << '\n';
	
	return 0;
}