2018年北京信息科技大学第十届程序设计竞赛暨ACM选拔赛

A PUBG <dijkstra>

边权不为1时最短路使用dijkstra。

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 110;
int g[N][N];
bool vis[N][N];
int dis[N][N];
int n, sx, sy, ex, ey;
int dir[4][2] = { -1, 0, 1, 0, 0, -1, 0, 1 };

struct node
{
	int x, y, k;
	bool operator < (const node &o) const
	{
		return k > o.k;
	}
};
int dijkstra()
{
	memset(vis, 0, sizeof(vis));
	memset(dis, 0x3f, sizeof(dis));
	dis[sx][sy] = 0;
	priority_queue<node> pq;
	pq.push({ sx, sy, 0 });
	while (!pq.empty())
	{
		int x = pq.top().x, y = pq.top().y, k = pq.top().k; pq.pop();
		vis[x][y] = 1;
		for (int i = 0; i < 4; ++i)
		{
			int xx = x + dir[i][0], yy = y + dir[i][1];
			if (xx == ex && yy == ey)
				return k;
			if (xx >= 1 && xx <= n && yy >= 1 && yy <= n && !vis[xx][yy] && dis[xx][yy] > k + g[xx][yy])
			{
				dis[xx][yy] = k + g[xx][yy];
				pq.push({ xx, yy, dis[xx][yy] });
			}
		}
	}
	return -1;
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	while (cin >> n)
	{
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)
			{
				scanf("%d", &g[i][j]);
				if (g[i][j] == -1)
					sx = i, sy = j;
				else if (g[i][j] == -2)
					ex = i, ey = j;
			}
		int res = dijkstra();
		cout << res << endl;
	}

	return 0;
}

B precise math function

有精度损失，不要用变量存直接printf输出。

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

int main()
{
#ifdef LOCAL
	//freopen("C:/input.txt", "r", stdin);
#endif
	int T;
	cin >> T;
	while (T--)
	{
		int n, x;
		cin >> n >> x;
		char s[] = "%.0f\n";
		s[2] = '0' + x;
		printf(s, pow(n, acos(-1)));
	}

	return 0;
}

D 打篮球 <模拟>

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
int a[110];
int n;

void solve()
{
	for (int i = 0; i < n; ++i)
		scanf("%d", &a[i]);
	int b[2] = { 1, 2 };
	for (int i = 0; i < n; ++i)
	{
		if (b[0] != a[i] && b[1] != a[i])
		{
			cout << "NO" << endl;
			return;
		}
		for (int j = 1; j <= 3; ++j)
			if (j != b[0] && j != b[1])
			{
				b[1] = j;
				b[0] = a[i];
				break;
			}
	}
	cout << "YES" << endl;
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	while (cin >> n)
		solve();

	return 0;
}

E 233 <高精度>

import java.util.*;
import java.math.*;
public class Main
{
	public static void main(String args[])
	{
		Scanner sc = new Scanner(System.in);
		int T;
		T = sc.nextInt();
		for (int ti = 1; ti <= T; ++ti)
		{
			BigInteger a, b;
			a = sc.nextBigInteger();
			b = sc.nextBigInteger();
			System.out.println(a.multiply(b));
		}
	}
}

F 扫雷 <模拟>

要了解扫雷整块扩散的规则，沿着一个方向扩散的时候，遇到第一个非0块则不再继续向前。
每次扫描8个方向计算当前块数字。

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 510;
int dir[8][2] = { -1, 0, 1, 0, 0, -1, 0, 1, -1, -1, -1, 1, 1, -1, 1, 1 };
int n, m, k;
char g[N][N], p[N][N];
int a[N], b[N];

void DFS(int x, int y, int z)
{
	if (p[x][y] != '.' || g[x][y] == '*')
		return;
	int cnt = 0;
	for (int i = 0; i < 8; ++i)
	{
		int xx = x + dir[i][0], yy = y + dir[i][1];
		if (xx >= 1 && xx <= n && yy >= 1 && yy <= m)
		{
			if (g[xx][yy] == '*')
				++cnt;
			else if (p[xx][yy] == '0')
				++z;
		}
	}
	if (cnt && !z)
		return;
	p[x][y] = '0' + cnt;
	for (int i = 0; i < 4; ++i)
	{
		int xx = x + dir[i][0], yy = y + dir[i][1];
		if (xx >= 1 && xx <= n && yy >= 1 && yy <= m)
			DFS(xx, yy, 0);
	}
}
void solve()
{
	for (int i = 1; i <= k; ++i)
	{
		int x = a[i], y = b[i];
		if (g[x][y] == '*')
		{
			printf("Game over in step %d\n", i);
			return;
		}
		DFS(x, y, 1);
		/*
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
				putchar(p[i][j]);
			putchar('\n');
		}
		cout << endl;
		*/
	}
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
			putchar(p[i][j]);
		putchar('\n');
	}
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	int T;
	cin >> T;
	while (T--)
	{
		memset(p, '.', sizeof(p));
		cin >> n >> m >> k;
		for (int i = 1; i <= n; ++i)
			scanf("%s", g[i] + 1);
		for (int i = 1; i <= k; ++i)
			scanf("%d%d", &a[i], &b[i]);
		solve();
	}

	return 0;
}

G 火车上的2连座

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
char s[1010][100];
int n;

void solve()
{
	for (int i = 1; i <= n; ++i)
		scanf("%s", s[i]);
	for (int i = 1; i <= n; ++i)
	{
		if (s[i][0] == 'O' && s[i][1] == 'O')
		{
			s[i][0] = s[i][1] = '+';
			cout << "YES" << endl;
			for (int j = 1; j <= n; ++j)
				printf("%s\n", s[j]);
			return;
		}
		if (s[i][3] == 'O' && s[i][4] == 'O')
		{
			s[i][3] = s[i][4] = '+';
			cout << "YES" << endl;
			for (int j = 1; j <= n; ++j)
				printf("%s\n", s[j]);
			return;
		}
	}
	cout << "NO" << endl;
}
int main()
{
#ifdef LOCAL
	//freopen("C:/input.txt", "r", stdin);
#endif
	while (cin >> n)
		solve();

	return 0;
}

H 程序员的好印象

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 110;
int a[N], f[2][N];

int main()
{
#ifdef LOCAL
	//freopen("C:/input.txt", "r", stdin);
#endif
	int n;
	while (cin >> n)
	{
		memset(f, 0, sizeof(f));
		for (int i = 1; i <= n; ++i)
			scanf("%d", &a[i]);
		int ans = 0;
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 0; j < i; ++j)
			{
				f[0][i] = max(f[0][i], f[0][j]);
				f[1][i] = max({ f[0][i], f[0][j], f[1][j] });
			}
			ans = max(ans, ++f[a[i]][i]);
		}
		cout << ans << endl;
	}

	return 0;
}