刷题日志-USACO Section 2全套

[P1457 USACO2.1]城堡 The Castle 思维难度不高，难度居然是蓝，被某谷赋虚高了，主要是考码力，地图需要在每个点上统计东南西北四处是否有墙，好在没有冲突的情况，无需分类讨论（实际上我也不会/qwq）。剩下的就是DFS洪水填充，统计一下就可以了，反正范围只到50。

#include 
#include 
#include 
// #include 

using namespace std;

const int MAXN = 55;
const int dir[5][2] = {
   
	{
   0, 0}, {
   0, -1}, {
   -1, 0}, {
   0, 1}, {
   1, 0}
};
int n, m, cnt, maxn, total;
bool a[MAXN][MAXN][4], vis[MAXN][MAXN];

void Judge(int x, int y, int t) {
   
//	1: 在西面有墙(1)
//	2: 在北面有墙(2)
//	4: 在东面有墙(3) 
//	8: 在南面有墙(4)
    // 这段写累死了
	switch(t) {
   
		case 1:
			a[x][y][1] = true;
			break;
		case 2:
			a[x][y][2] = true;
			break;
		case 3:
			a[x][y][1] = a[x][y][2] = true;
			break;
		case 4: 
			a[x][y][3] = true;
			break;
		case 5: 
			a[x][y][1] = a[x][y][3] = true;
			break;
		case 6: 
			a[x][y][2] = a[x][y][3] = true;
			break;
		case 7: 
			a[x][y][1] = a[x][y][2] = a[x][y][3] = true;
			break;
		case 8: 
			a[x][y][4] = true;
			break;
		case 9: 
			a[x][y][1] = a[x][y][4] = true;
			break;
		case 10: 
			a[x][y][2] = a[x][y][4] = true;
			break;
		case 11: 
			a[x][y][1] = a[x][y][2] = a[x][y][4] = true;
			break;
		case 12: 
			a[x][y][3] = a[x][y][4] = true;
			break;
		case 13: 
			a[x][y][1] = a[x][y][3] = a[x][y][4] = true;
			break;
		case 14: 
			a[x][y][2] = a[x][y][3] = a[x][y][4] = true;
			break;
		case 15: 
			a[x][y][1] = a[x][y][2] = a[x][y][3] = a[x][y][4] = true;
			break;
		default:
			break;
	}
	// cout << t << " " << a[x][y][1] << " " << a[x][y][2] << " " << a[x][y][3] << " " << a[x][y][4] << endl; 
}

bool valid(int x, int y) {
   
	return x >= 1 && y >= 1 && x <= n && y <= m;
}
void dfs(int x, int y) {
   
	// cout << x << " " << y << endl;
	vis[x][y] = true;
	cnt ++;
	for(int i=1; i<=4; i++) {
   
		int nx = x + dir[i][0], ny = y + dir[i][1];
		if(valid(nx, ny) && !a[x][y][i] && !vis[nx][ny]) {
   
			dfs(nx, ny);
		}
	}
	// vis[x][y] = false; 洪水填充 
}
int maxt, posx, posy;
char c;

int main() {
   
	cin >> m >> n;
	for(int i=1; i<=n; i++) {
   
		for(int j=1; j<=m; j++) {
   
			int t;
			cin >> t;
			Judge(i, j, t);
		}
	}
	memset(vis, false, sizeof(vis));
	/*
	for(int i=1; i<=n; i++) {
		for(int j=1; j<=m; j++) {
			cout << i << " " << j << " " << a[i][j][1] << " " << a[i][j][2] << " " << a[i][j][3] << " " << a[i][j][4] << endl; 
		}
	} 
	*/
	for(int i=1; i<=n; i++) {
   
		for(int j=1; j<=m; j++) {
   
			if(!vis[i][j]) {
   
				// cout << i << " " << j << endl;
				cnt = 0;
				dfs(i, j);
				total ++;
				maxn = max(maxn, cnt);
				// cout << endl;
			}
		}
	}
	cout << total << endl << maxn << endl;
	// system("pause");
	for(int j=1; j<=m; j++) {
   
		for(int i=n; i>=1; i--) {
   
			if(a[i][j][2]) {
   
				memset(vis, false, sizeof(vis));
				a[i][j][2] = false;
				cnt = 0;
				dfs(i, j);
				a[i][j][2] = true;
				if(cnt > maxt) {
   
					maxt = cnt;
					posx = i, posy = j;
					c = 'N';
				}
			}
			if(a[i][j][3]) {
   
				memset(vis, false, sizeof(vis));
				a[i][j][3] = false;
				cnt = 0;
				dfs(i, j);
				a[i][j][3] = true;
				if(cnt > maxt) {
   
					maxt = cnt;
					posx = i, posy = j;
					c = 'E';
				}
			}
		}
	}
	cout << maxt << endl << posx << " " << posy << " " << c;
	return 0;
}

[P1458 USACO2.1]顺序的分数 Ordered Fractions 按部就班模拟即可，唯一的坑点就是0和某个数的最大公约数需要特判

#include 
#include 
 
using namespace std;

int n, m;
struct node {
   
	int p, q;
	double c;
} a[165*165];

int gcd(int x, int y) {
   
	return x%y==0?y:gcd(y,x%y);
}

bool cmp(node x, node y) {
   
	return x.c < y.c;
}

int main() {
   
	cin >> n;
	for(int i=1; i<=n; i++) {
   
		for(int j=0; j<=i; j++) {
   
			int k;
			if(j == 0) k = i;
			else k = gcd(i, j);
			if(k == 1) {
   
				m ++;