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【补题记录】ICPC2023 Jinan

时间：2024-01-21 20:33:25浏览次数：40  
标签：ICPC2023 cnt ch return int Jinan read 补题 col

【补题记录】ICPC2023 Jinan

Contest Link: https://qoj.ac/contest/1472.

Problems: https://sua.ac/wiki/2023-icpc-jinan/contest-zh.pdf.

Solution: https://qoj.ac/download.php?type=attachments&id=1472&r=1.

A. Many Many Heads

const int N = 1e5 + 10;
int T;
string s;

void solve(){
	read(T);
	while(T--){
		read(s);
		int n = s.size();
		#define cg(x) (x=='['||x==']'?1:0)
		bool flg = 1;
		for(int i = 0; i + 2 < n; ++ i){
			if(cg(s[i]) == cg(s[i+1]) && cg(s[i+1]) == cg(s[i+2])){
				flg = 0;
				break;
			}
		}
		int cnt = 0;
		for(int i = 0; i + 1 < n; ++ i){
			if(cg(s[i]) == cg(s[i+1])){
				++ cnt;
			}
		}
		if(cnt > 2){
			flg = 0;
		}
		println_cstr((flg ? "Yes" : "No"));
	}
}

B. Graph Partitioning 2

C. Turn on the Light 2

D. Largest Digit

int clc(int x){
	int mx = -1;
	while(x){
		mx = max(mx, x%10);
		x /= 10;
	}
	return mx;
}

void solve(){
	int T, a, b, x, y;
	read(T);
	while(T--){
		read(a, b, x, y);
		if(b - a >= 10 || y - x >= 10){
			println(9);
		} else {
			int mx = -1;
			for(int i = a; i <= b; ++ i){
				for(int j = x; j <= y; ++ j){
					mx = max(mx, clc(i+j));
				}
			}
			println(mx);
		}
	}
}

E. I Just Want... One More...

F. Say Hello to the Future

G. Gifts from Knowledge

const int N = 3e6 + 10;
int T, n, m, pool[N*2], col[N];
string s;
bool flg = false;
vector<pair<int, int> > g[N];
const ll P = 1e9 + 7;
ll pw2[N];

void dfs(int x, int cl){
	col[x] = cl;
	for(auto i : g[x]){
		int y = i.first, z = i.second;
		if(col[y] != -1 && (col[y] ^ z ^ col[x])){
			flg = true;
		}
		if(col[y] == -1){
			dfs(y, cl ^ z);
		}
	}
}

void solve(){
	memset(col, -1, sizeof(col));
	pw2[0] = 1;
	for(int i = 1; i < N; ++ i){
		pw2[i] = pw2[i-1] * 2 % P;
	}
	read(T);
	while(T--){
		read(n, m);
		int (&a)[n+5][m+5] = decltype(a)(pool);
		for(int i = 1; i <= n; ++ i){
			read(s);
			for(int j = 1; j <= m; ++ j){
				a[i][j] = s[j-1] - '0';
			}
		}
		flg = false;
		for(int i = 1, j = m; i <= j; ++ i, -- j){
			int cnt = 0, x = 0, y = 0;
			for(int k = 1; k <= n; ++ k){
				cnt += a[k][i] + a[k][j];
				if(a[k][i] == 1 && a[k][j] != 1){
					if(x){
						if(y == i){
							g[k].emplace_back(x, 1);
							g[x].emplace_back(k, 1);
						} else {
							g[k].emplace_back(x, 0);
							g[x].emplace_back(k, 0);
						}
					} else {
						x = k, y = i;
					}
				} else if(a[k][j] == 1 && a[k][i] != 1){
					if(x){
						if(y == j){
							g[k].emplace_back(x, 1);
							g[x].emplace_back(k, 1);
						} else {
							g[k].emplace_back(x, 0);
							g[x].emplace_back(k, 0);
						}
					} else {
						x = k, y = j;
					}
				}
			}
			if(cnt > 2){
				flg = true;
				break;
			}
		}
		if(flg){
			println(0);
		} else {
			int cc = 0;
			for(int i = 1; i <= n; ++ i){
				if(col[i] == -1){
					dfs(i, 0);
					++ cc;
				}
			}
			if(flg){
				println(0);
			} else {
				println(pw2[cc]);
			}
		}
		for(int i = 1; i <= n; ++ i){
			col[i] = -1;
			vector<pair<int, int> > ().swap(g[i]);
		}
	}
}

H. Basic Substring Structure

I. Strange Sorting

const int N = 110;
int T, n, a[N], b[N];

void solve(){
	read(T);
	while(T--){
		read(n);
		for(int i = 1; i <= n; ++ i){
			read(a[i]);
		}
		memcpy(b, a, sizeof(a));
		int ans = 0;
		for(int i = 1; i <= n; ++ i){
			if(a[i] != i){
				for(int j = n; j > i; -- j){
					if(a[i] > a[j]){
						++ ans;
						sort(a + i, a + j + 1);
						break;
					}
				}
			}
		}
		println(ans);
		memcpy(a, b, sizeof(b));
		for(int i = 1; i <= n; ++ i){
			if(a[i] != i){
				for(int j = n; j > i; -- j){
					if(a[i] > a[j]){
						println(i, j);
						sort(a + i, a + j + 1);
						break;
					}
				}
			}
		}
	}
}

J. Computational Intelligence

K. Rainbow Subarray

const int N = 1e6 + 10;
int T, n, a[N];
ll k, bit[N], bb[N], b[N];

void add(int x, ll v){
	while(x <= n){
		bit[x] += v;
		x += x & -x;
	}
}
void addd(int x, ll v){
	while(x <= n){
		bb[x] += v;
		x += x & -x;
	}
}
ll ask(int x){
	ll res = 0;
	while(x){
		res += bit[x];
		x -= x & -x;
	}
	return res;
}
ll askk(int x){
	ll res = 0;
	while(x){
		res += bb[x];
		x -= x & -x;
	}
	return res;
}

int ch[N][2], val[N], pri[N], siz[N], tot;
int root, x, y, z;

void update(int x){
	siz[x] = siz[ch[x][0]] + siz[ch[x][1]] + 1;
}
int newnode(int v){
	siz[++tot] = 1;
	val[tot] = v;
	pri[tot] = rand();
	return tot;
}
int merge(int x, int y){
	if(!x || !y){
		return x + y;
	}
	if(pri[x] < pri[y]){
		ch[x][1] = merge(ch[x][1], y);
		update(x);
		return x;
	} else {
		ch[y][0] = merge(x, ch[y][0]);
		update(y);
		return y;
	}
}
void split(int p, int k, int &x, int &y){
	if(!p){
		x = y = 0;
		return;
	}
	if(val[p] <= k){
		x = p;
		split(ch[p][1], k, ch[p][1], y);
	} else {
		y = p;
		split(ch[p][0], k, x, ch[p][0]);
	}
	update(p);
}
int kth(int p, int k){
	while(true){
		if(k <= siz[ch[p][0]]){
			p = ch[p][0];
		} else if(k == siz[ch[p][0]] + 1){
			return p;
		} else {
			k -= siz[ch[p][0]] + 1;
			p = ch[p][1];
		}
	}
}
void ins(int k){
	split(root, k, x, y);
	root = merge(merge(x, newnode(k)), y);
}
void del(int k){
	split(root, k, x, z);
	split(x, k-1, x, y);
	y = merge(ch[y][0], ch[y][1]);
	root = merge(merge(x, y), z);
}
int getval(int k){
	return val[kth(root, k)];
}

void solve(){
	read(T);
	while(T--){
		read(n, k);
		for(int i = 1; i <= n; ++ i){
			read(a[i]);
			a[i] -= i;
			b[i] = a[i];
		}
		sort(b + 1, b + n + 1);
		int m = unique(b + 1, b + n + 1) - b - 1;
		for(int i = 1; i <= n; ++ i){
			a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
		}
		int ans = 1;
		add(a[1], b[a[1]]);
		addd(a[1], 1);
		ins(a[1]);
		for(int l = 1, r = 1; l <= n; ++ l){
			r = max(r, l-1);
			while(r < n){
				add(a[r+1], b[a[r+1]]);
				addd(a[r+1], 1);
				ins(a[r+1]);
				int rr = ((r+1) - l + 1) / 2 + 1;
				int mid = getval(rr);
				ll val = b[mid] * askk(mid-1) - ask(mid-1);
				val += ask(m) - ask(mid) - b[mid] * (askk(m) - askk(mid));
				if(abs(val) <= k){
					ans = max(ans, (r+1) - l + 1);
					++ r;
				} else {
					add(a[r+1], -b[a[r+1]]);
					addd(a[r+1], -1);
					del(a[r+1]);
					break;
				}
			}
			add(a[l], -b[a[l]]);
			addd(a[l], -1);
			del(a[l]);
		}
		println(ans);
		for(int i = 1; i <= n; ++ i){
			bit[i] = bb[i] = b[i] = 0;
		}
		for(int i = 1; i <= tot; ++ i){
			ch[i][0] = ch[i][1] = siz[i] = val[i] = pri[i] = 0;
		}
		tot = root = 0;
	}
}

L. Ticket to Ride

M. Almost Convex

const int N = 4e3 + 10;
const double eps = 1e-8;
int n, x[N], y[N], is[N], id[N];

int is0(double x){return (fabs(x)<eps?0:(x<0)?-1:1);}
double at[N];
struct Point{
	double x,y;
	int id;
	Point(){}
	Point(double x,double y):x(x),y(y){}
	Point operator+(Point b){return Point(x+b.x,y+b.y);}
	Point operator-(Point b){return Point(x-b.x,y-b.y);}
	bool operator==(Point b){return is0(x-b.x)==0&&is0(y-b.y)==0;}
	bool operator<(Point b){return is0(x-b.x)<0||(is0(x-b.x)==0&&is0(y-b.y)<0);}
}p[N],ch[N];
typedef Point Vector;
typedef Point point;
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}
double Dist(Point a,Point b){return hypot(a.x-b.x,a.y-b.y);}
int Andrew(int n){
	int v=0,k;
	sort(p,p+n);
	n=unique(p,p+n)-p;
	for(int i=0; i<n; ++i){
		while(v>1 && is0(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-2]))<1) --v;
		ch[v++]=p[i];
	}
	for(int i=n-2,k=v; i>=0; --i){
		while(v>k && is0(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-2]))<1) --v;
		ch[v++]=p[i];
	}
	return (n>1?v-1:v);
}

bool cmp1(point x, point y){
	if(at[x.id] != at[y.id]){
		return at[x.id] < at[y.id];
	}
	return x.x < y.x;
}
int qdr(point x){
	if(x.x > 0 && x.y >= 0) return 1;
	if(x.x <= 0 && x.y > 0) return 2;
	if(x.x < 0 && x.y <= 0) return 3;
	if(x.x >= 0 && x.y < 0) return 4;
}
bool cmp(point x, point y){
	if(qdr(x) == qdr(y)){
		return cmp1(x, y);
	} else {
		return qdr(x) < qdr(y);
	}
}

int v;

int calc(int k){
	int cnt = 0, ans = 0;
	for(int i = 0; i < n; ++ i){
		if(i == k){
			continue;
		}
		ch[++cnt].x = x[i] - x[k] + eps;
		ch[cnt].y = y[i] - y[k] + eps;
		ch[cnt].id = i;
		at[i] = atan2(ch[cnt].y, ch[cnt].x);
	}
	sort(ch + 1, ch + cnt + 1, cmp);
	ch[cnt+1] = ch[1];
	for(int i = 1; i <= cnt; ++ i){
		if(is[ch[i].id] && is[ch[i+1].id]){
			if(abs(id[ch[i].id] - id[ch[i+1].id]) == 1){
				++ ans;
			} else if(id[ch[i].id] == v && id[ch[i+1].id] == 1){
				++ ans;
			} else if(id[ch[i].id] == 1 && id[ch[i+1].id] == v){
				++ ans;
			}
		}
	}
	return ans;
}

void solve(){
	read(n);
	for(int i = 0; i < n; ++ i){
		read(x[i], y[i]);
		p[i].x = x[i];
		p[i].y = y[i];
		p[i].id = i;
	}
	v = Andrew(n);
	for(int i = 0; i < v; ++ i){
		is[ch[i].id] = 1;
		id[ch[i].id] = i + 1;
	}
	int ans = 1;
	for(int i = 0; i < n; ++ i){
		if(!is[i]){
			ans += calc(i);
		}
	}
	println(ans);
}

标签：ICPC2023,cnt,ch,return,int,Jinan,read,补题,col
From： https://www.cnblogs.com/KiharaTouma/p/17978297/ICPC2023-Jinan

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