AtCoder Beginner Contest 197（Sponsored by Panasonic） Person Editorial

A - Rotate

先输出第二和第三个字符，然后再输出第一个字符即可

B - Visibility

以 $(x,y)$ 作为起点向4个方向探索不是 # 的点，注意一下会在$(x,y)$重复计算 $3$ 次，所以要 cnt - 3

int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int H, W, X, Y;
    cin >> H >> W >> X >> Y;
    X -= 1, Y -= 1;
    vector<string> s(H);
    for (int i = 0; i < H; ++i) cin >> s[i];
    int cnt = 0;
    // 向 4 个方向探索
    for (int i = X; i < H and s[i][Y] != '#'; ++i) cnt++;
    for (int i = X; i >= 0 and s[i][Y] != '#'; --i) cnt++;
    for (int i = Y; i < W and s[X][i] != '#'; ++i) cnt++;
    for (int i = Y; i >= 0 and s[X][i] != '#'; --i) cnt++;
    cout << cnt - 3 << "\n";
    return 0;
}

C - ORXOR Editorial

Good，位运算典型题

int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int n;
    cin >> n;
    vector<int> a(n);
    for (int& x : a) cin >> x;
    int ans = INT_MAX;
    for (int i = 0; i < (1 << (n - 1)); ++i) {
        int xored = 0, ored = 0;
        for (int j = 0; j <= n; ++j) {
            if (j < n) ored |= a[j];
            if (j == n || (i >> j & 1)) xored ^= ored, ored = 0;
        }
        ans = min(ans, xored);
    }
    cout << ans << "\n";
    return 0;
}

D - Opposite

// C++似乎内置了 PI 也可以不定义 M_PI
#define M_PI 3.14159265358979323846
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int N;
    double xx, yy, x, y;
    cin >> N >> xx >> yy >> x >> y;
    double a = (xx + x) / 2, b = (yy + y) / 2;
    xx -= a, yy -= b;
    double m = xx * cos(2 * M_PI / N) - yy * sin(2 * M_PI / N);
    double n = xx * sin(2 * M_PI / N) + yy * cos(2 * M_PI / N);
    cout << fixed << setprecision(12) << m + a << "\n" << n + b;
    return 0;
}

E - Traveler

很好的处理方法

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 2e5 + 7;
vector<int> v[N];
ll l[N], r[N];
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int n;
    cin >> n;
    for (int i = 0, x, c; i < n; ++i) {
        cin >> x >> c;
        v[c].push_back(x);
    }
    v[0].push_back(0);
    v[n + 1].push_back(0);

    // int ans = 0;
    for (int i = 1, j = 0; i <= n + 1; ++i) {
        if (v[i].empty()) continue;
        sort(v[i].begin(), v[i].end());
        int lx = v[i].front(), rx = v[i].back();
        int ly = v[j].front(), ry = v[j].back();
        // cout << lx << " " << rx << "\n";
        l[i] = min(abs(rx - ly) + l[j], abs(rx - ry) + r[j]) + rx - lx;
        r[i] = min(abs(lx - ly) + l[j], abs(lx - ry) + r[j]) + rx - lx;
        // ans = min(l[i], r[i]);
        j = i;
    }

    cout << min(l[n + 1], r[n + 1]) << "\n";
    return 0;
}

F - Construct a Palindrome

#include <bits/stdc++.h>
using namespace std;
const int N = 1002;
int n, m;
char s[2];
struct node {
    int s, t, sp;
} r;
queue<node> q;
vector<int> a[N][26];
int ans = 1e9, vis[N][N];
//把边看成点
//对于两对边(a,b)  (c,d)
//如果(a,c) 和 (b,c)之间存在边 而且边上的字母相同的话
//那么这两个边变成的点就可以联通
int bfs() {
    while (!q.empty()) {
        r = q.front();
        q.pop();
        if (r.sp == ans) return ans << 1;
        for (int i = 0; i < 26; ++i)
            for (int j = 0; j < a[r.s][i].size(); ++j)
                for (int k = 0; k < a[r.t][i].size(); ++k) {
                    int ns = a[r.s][i][j];
                    int nt = a[r.t][i][k];
                    if (ns == r.t || nt == r.s) return r.sp << 1 | 1;
                    if (ns == nt) ans = r.sp + 1;
                    if (vis[ns][nt]) continue;
                    vis[ns][nt] = 1;
                    q.push((node){ns, nt, r.sp + 1});
                }
    }
    return -1;
}
int main() {
    scanf("%d%d", &n, &m);
    for (int x, y; m; --m) {
        scanf("%d%d%s", &x, &y, s);
        a[x][*s - 'a'].push_back(y);
        a[y][*s - 'a'].push_back(x);
    }
    vis[1][n] = 1;
    q.push((node){1, n, 0});
    printf("%d", bfs());
}