2020ICPC·小米 网络选拔赛第一场

2020ICPC·小米 网络选拔赛第一场

C-Smart Browser

#include <string>
#include <iostream>

std::string s;

void solve() {
	std::cin >> s;
	int ans = 0, t = s[0] == 'w';
	for (int i = 1; i < s.size(); i++) {
		if (s[i] == 'w') {
			t++;
		} else {
			if (t != 0) {
				ans += 2 * t - 1;
				t = 0;
			}
		}
	}
	if (t != 0) {
		ans += 2 * t - 1;
	}
	printf("%d\n" ,ans);
}

int main() {
	solve();
	return 0;
}

---

I-Walking Machine

思路：

一道模拟+DFS。对于每个点检查它的四个方向上的点能不能到达这个点，可以的话就DFS。

代码中处理了大量重复片段，一来减小了代码量，二来方便查错，具体细节在代码中体现。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>

#define pii pair<int, int>
#define mp(a, b) make_pair(a, b)
#define fr first
#define sc sesond

const int Maxn = 1005;
const int dir[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; // W S A D

int a[Maxn][Maxn], n, m, ans = 0;
bool vis[Maxn][Maxn];

inline int read() {
	char ch, tar[] = "WSAD";
	while ((ch = getchar()) != EOF) {
		for (int i = 0; i < 4; i++) {
			if (ch == tar[i]) {
				return i;
			}
		}
	}
	return -1;
}

std::pii getDest(int x, int y) {
	int d = a[x][y];
	int xx = x + dir[d][0];
	int yy = y + dir[d][1];
	return std::mp(xx, yy);
}

void run(int x, int y, int xx, int yy);

void dfs(int x, int y) {
	for (int i = 0; i < 4; i++) {
		int xx = x + dir[i][0];
		int yy = y + dir[i][1];
		if (xx >= 1 && yy >= 1 && xx <= n && yy <= m) {
			run(x, y, xx, yy);
		}
	}
}

void run(int x, int y, int xx, int yy) {
	if (std::mp(x, y) == getDest(xx, yy) && !vis[xx][yy]) {
		vis[xx][yy] = true;
		ans++;
		dfs(xx, yy);
	}
}

void solve() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			a[i][j] = read();
		}
	}
	for (int i = 1; i <= n; i++) {
		run(i, 0, i, 1);
		run(i, m + 1, i, m);
	}
	for (int i = 1; i <= m; i++) {
		run(0, i, 1, i);
		run(n + 1, i, n, i);
	}
	printf("%d\n", ans);
}

int main() {
	solve();
	return 0;
}

---

A-Intelligent Warehouse

思路：

记忆化搜索。对于每个存在的数字进行一次搜索，而一些数字可能在搜索之前的数字时就被搜索过，那么直接返回之前搜索过的结果即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

const int Maxn = 10000005;

int in[Maxn], a[Maxn], cnt[Maxn], maxx;

int dfs(int cur) {
	if (cnt[cur] != 0) {
		return cnt[cur];
	}
	int res = a[cur]; // 至少应该包括他自己
	for (int i = 2; i * cur <= maxx; i++) {
		if (a[i * cur]) {
			res = std::max(res, a[cur] + dfs(i * cur));
		}
	}
	cnt[cur] = res;
	return res;
}

void solve() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", in + i);
		maxx = std::max(maxx, in[i]);
		a[in[i]]++;
	}
	int ans = 0;
	for (int i = 0; i < n; i++) {
		ans = std::max(ans, dfs(in[i]));
	}
	printf("%d\n", ans);
}

int main() {
	solve();
	return 0;
}