2018-2019 ACM-ICPC, Asia Jiaozuo Regional Contest[6/12]

原创 于 2019-03-18 22:34:55 发布 · 314 阅读 · 0 ·
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博客分享了A - I几道算法题的解题情况。C题需对x和y分别排序并维护合并区间，用四个ans维护答案；D题pypy小数是float，交Python可过，还提供C++代码；E题待补；F题用bfs；I题贪心取两边。

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A

温暖签到

#include <bits/stdc++.h>

using namespace std;

int main() {
		int n;
		cin >> n;
		for (int i = 1; i <= n; ++i) {
				int x, ans = 0;
				for (int j = 1; j <= 4; ++j) {
						cin >> x;
						if (x > 0)
								++ans;
				}
				if (ans == 0) 
						puts("Typically Otaku");
				if (ans == 1) 
						puts("Eye-opener");
				if (ans == 2) 
						puts("Young Traveller");
				if (ans == 3)
						puts("Excellent Traveller");
				if (ans == 4)
						puts("Contemporary Xu Xiake");
		}
}

B

待补

C

upsolved
对x和y分别排序然后维护左边和右边分别被合并的区间，因为答案只可能出现在角落，所以就可以四个ans来维护(比赛的时候没看到每行每列只有一个这句话写死我了)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

namespace fastIO{  
    #define BUF_SIZE 100000  
    #define OUT_SIZE 100000  
    //fread->read  
    bool IOerror=0;  
    inline char nc(){  
		static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;  
		if (p1==pend){  
			p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);  
			if (pend==p1){IOerror=1;return -1;}  
			//{printf("IO error!\n");system("pause");for (;;);exit(0);}  
		}  
		return *p1++;  
	}  
	inline bool blank(char ch){return ch==32||ch==10||ch==13||ch==9;}  
	inline bool enter(char ch){return ch==10||ch==13;}
	inline void read(int &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(ll &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(double &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (ch==46){  
			double tmp=1; ch=nc();  
			for (;ch>=48&&ch<=57;ch=nc())tmp/=10.0,x+=tmp*(ch-48);  
		}  
		if (sign)x=-x;  
	}  
	inline void read(char *s){  
		char ch=nc();  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;  
		*s=0;  
	}  
	inline void readln(char *s) {
		char ch=nc();
		for (;blank(ch);ch=nc());
		if(IOerror)return;
		for(;!enter(ch)&&!IOerror;ch=nc())*s++=ch;
		*s=0;
	}
	inline void read(char &c){  
		for (c=nc();blank(c);c=nc());  
		if (IOerror){c=-1;return;}  
	} 
#undef OUT_SIZE  
#undef BUF_SIZE  
}
using fastIO::read;

int const N = 300005;

struct node {
	int x, y, id;
};
vector<node> detx, dety;
int ans1, ans2, ans3, ans4;
int lx, rx, ly, ry;
int xmi, xma, ymi, yma;
int dx, dy;
int mark[N];
int n, m;

void setpn(node a, int val) {
	if (mark[a.id] == 0) {
		mark[a.id] = val;
		return;
	}
	int& t = mark[a.id];
	if (t == 1) {
		if (val == 3)
			++ans1;
		if (val == 4)
			++ans3;
	}
	if (t == 2) {
		if (val == 3)
			++ans2;
		if (val == 4)
			++ans4;
		if (val == 1)
			t = val;
	}
	if (t == 3) {
		if (val == 1) 
			++ans1;
		if (val == 2)
			++ans2;
	}
	if (t == 4) {
		if (val == 1)
			++ans3;
		if (val == 2)
			++ans4;
		if (val == 3)
			t = val;
	}
}

void init(vector<node>& t) {
	for (int i = 0; i < n; ++i)
		mark[i] = 0;
	ans1 = ans2 = ans3 = ans4 = 0;
	dx = dy = 0;
	detx = t;
	dety = t;
	sort(detx.begin(), detx.end(), [] (node const& a, node const& b) {
			return a.x < b.x;
			});
	sort(dety.begin(), dety.end(), [] (node const& a, node const& b) {
			return a.y < b.y;
			});
	lx = 0, rx = n - 1;
	setpn(detx[0], 3);
	setpn(detx[n - 1], 4);
	xmi = detx[0].x;
	xma = detx[n - 1].x;

	ly = 0, ry = n - 1;
	setpn(dety[0], 1);
	setpn(dety[n - 1], 2);
	ymi = dety[0].y;
	yma = dety[n - 1].y;
}

void moveL(int step) {
	dy -= step;
	ymi = max(1, ymi - step);
	yma = max(1, yma - step);
	while (ly < ry && dety[ly + 1].y + dy <= ymi) {
		++ly;
		if (ly == ry) 
			return;
		setpn(dety[ly], 1);
	}
}

void moveR(int step) {
	dy += step;
	ymi = min(n, ymi + step);
	yma = min(n, yma + step);
	while (ly < ry && dety[ry - 1].y + dy >= yma) {
		--ry;
		if (ly == ry) 
			return;
		setpn(dety[ry], 2);
	}
}

void moveU(int step) {
	dx -= step;
	xmi = max(1, xmi - step);
	xma = max(1, xma - step);
	while (lx < rx && detx[lx + 1].x + dx <= xmi) {
		++lx;
		if (lx == rx) 
			return;
		setpn(detx[lx], 3);
	}
}

void moveD(int step) {
	dx += step;
	xmi = min(n, xmi + step);
	xma = min(n, xma + step);
	while (lx < rx && detx[rx - 1].x + dx >= xma) {
		--rx;
		if (lx == rx) 
			return;
		setpn(detx[rx], 4);
	}
}

int query(int const& x, int& mi, int& ma, int d) {
	if (x + d <= mi)
		return mi;
	if (x + d >= ma)
		return ma;
	return x + d;
}

long long C(int x) {
	return (long long) x * (x - 1) / 2;
}

long long QueryAns() {
	if (n == 1)
		return 0;
	if (xmi == xma && ymi == yma) 
		return C(ans1 + ans2 + ans3 + ans4);
	if (xmi == xma) 
		return C(ans1 + ans3) + C(ans2 + ans4);
	if (ymi == yma) 
		return C(ans1 + ans2) + C(ans3 + ans4);
	return C(ans1) + C(ans2) + C(ans3) + C(ans4);
}

int main() {
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
	ios::sync_with_stdio(0);
	int T;
	read(T);
	while (T--) {
		read(n); 
		read(m);
		auto a = vector<node>(n);
		for (int i = 0; i < n; ++i) {
			read(a[i].x);
			read(a[i].y);
			a[i].id = i;
		}
		init(a);
		/////////////////////
			// cerr << "MARK : ";
			// for (int i = 0; i < n; ++i)
			//     cerr << mark[i] << ' ';
			// cerr << '\n';

		for (int o = 0; o < m; ++o) {
			char op;
			int x;
			read(op);
			if (op != '!')
				read(x);
			switch(op) {
				case 'L': 
					moveL(x);
					break;
				case 'R':
					moveR(x);
					break;
				case 'U':
					moveU(x);
					break;
				case 'D':
					moveD(x);
					break;
				case '?': {
					int px = query(a[x - 1].x, xmi, xma, dx);
					int py = query(a[x - 1].y, ymi, yma, dy);
					cout << px << ' ' << py << '\n';
					}
					break;
				default :
					cout << QueryAns() << '\n';
			}

		}
	}
}

D

pypy的小数居然是float，交Python就能过，这里提供相同的C++代码

#include <bits/stdc++.h>

using namespace std;

double const PI = acos(-1.0);

double sqr(double x) {
		return x * x;
}

double calc(double a, double b, double r, double th) {
		th = th / 180.0 * PI;
		double l = sqrt(sqr(a + r) + sqr(b));
		double th0 = atan(b / (a + r));
		return l * cos(max(th0 - th, 0.0)) - r;
}

int main() {
		int T;
		cin >> T;
		while (T--) {
				double a, b, r, th;
				cin >> a >> b >> r >> th;
				double ans = calc(a, b, r, th);
				cout << fixed << setprecision(12) << ans << '\n';
		}
}

E

from math import gcd

def primes(mx):
    ret = []
    for i in range(2, mx):
        if all(i % p != 0 for p in ret):
            ret.append(i)
    return ret

def select(ps, n):
    cur, xs = 1, []
    for p in ps:
        if cur * p > n:
            return cur, xs
        cur *= p
        xs.append(p)

def calc(n, ps):
    a, qs = select(ps, n)
    b = 1
    for q in qs:
        b *= q + 1
    g = gcd(a, b)
    return a // g, b // g

def main():
    ps = primes(300)

    for _ in range(int(input())):
        n = int(input())
        a, b = calc(n, ps)
        print(f'{a}/{b}')

main()

F

bfs

#include <iostream>
#include <string>
#include <queue>
#include <utility>
#include <algorithm>

using Edges = std::vector<std::vector<int>>;

int bfs(int n, const Edges &es, int ibeg, int iend) {
	const int INF = 1000 * 1000 + 10;
	auto q = std::queue<int>();
	auto dis = std::vector<int>(n, INF);
	q.push(ibeg);
	dis[ibeg] = 0;
	while(!q.empty()) {
		int c = q.front();
		q.pop();
		if(c == iend)
			return 1 + dis[c];
		for(int to: es[c])
			if(dis[to] == INF) {
				dis[to] = dis[c] + 1;
				q.push(to);
			}
	}
	return -1;
}

int main() {
	std::ios::sync_with_stdio(false);

	int t;
	std::cin >> t;
	while(t--) {
		int r, c;
		std::cin >> r >> c;
		std::cin.get(); // Eat space.
		int n = r * c, ibeg, iend;

		std::vector<std::string> inp;
		for(int i = 0; i < 4 * r + 3; ++i) {
			std::string s;
			std::getline(std::cin, s);
			inp.push_back(std::move(s));
		}

		auto es = Edges(n);

		auto adde = [&](int x1, int y1, int x2, int y2) {
			es[x1 * c + y1].push_back(x2 * c + y2);
			es[x2 * c + y2].push_back(x1 * c + y1);
		};

		for(int i = 0; i < r; ++i)
			for(int j = 0; j < c; ++j) {
				int x = (j & 1 ? 4 : 2) + i * 4,
					y = 4 + j * 6;

				if(inp[x][y] == 'S')
					ibeg = i * c + j;
				if(inp[x][y] == 'T')
					iend = i * c + j;

				if(inp[x + 2][y] == ' ')
					adde(i, j, i + 1, j);
				if(inp[x + 1][y + 3] == ' ') {
					if(j & 1)
						adde(i, j, i + 1, j + 1);
					else
						adde(i, j, i, j + 1);
				}
				if(inp[x - 1][y + 3] == ' ') {
					if(j & 1)
						adde(i, j, i, j + 1);
					else
						adde(i, j, i - 1, j + 1);
				}
			}
		
		int ans = bfs(n, es, ibeg, iend);
		std::cout << ans << '\n';

	}

    return 0;
}

I

贪心取两边即可

#include<iostream>
using namespace std;
long long a[100010];
int main()
{
		int t,n;
		cin>>t;
		while(t--)
		{
				long long ans=0,sum=0;
				cin>>n;
				for(int i=1;i<n;i++)cin>>a[i];
				for(int i=1;i<n;i++)a[i]+=a[i-1];
				for(int i=1;i<=n;i++)
				{
						if(i&1)ans+=sum;
						else sum+=a[n-i/2]-a[i/2-1],ans+=sum;
						cout<<ans;
						if(n!=i)cout<<" ";
				}
				cout<<endl;
		}
}
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