Round A 2021

比赛链接：https://codingcompetitions.withgoogle.com/kickstart/round/0000000000436140

K-Goodness String (5pts, 7pts)

判断字符不同值，计算得分至K的差值

int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int T;
    cin >> T;
    for (int i = 1; i <= T; ++i) {
        int n, k, cnt = 0;
        cin >> n >> k;
        string s;
        cin >> s;
        for (int i = 0; i < n / 2; ++i)
            if (s[i] != s[n - i - 1]) cnt++;
        // cout << cnt << "
";
        if (cnt == k) cnt = 0;
        else
            cnt = abs(k - cnt);
        cout << "Case #" << i << ": " << cnt << "
";
    }
    return 0;
}

L Shaped Plots (8pts, 12pts)

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int N, M, a[1005][1005], l[1005][1005], r[1005][1005], u[1005][1005],
    d[1005][1005];

int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int T;
    cin >> T;
    for (int ca = 1; ca <= T; ca++) {
        cin >> N >> M;
        for (int i = 1; i <= N; ++i)
            for (int j = 1; j <= M; ++j) cin >> a[i][j];

        memset(l, 0, sizeof(l)), memset(r, 0, sizeof(r));
        memset(u, 0, sizeof(u)), memset(d, 0, sizeof(d));
        for (int i = 1; i <= N; i++)
            for (int j = 1; j <= M; j++) {
                if (a[i][j])
                    l[i][j] = l[i][j - 1] + 1, u[i][j] = u[i - 1][j] + 1;
            }

        int ret = 0;
        for (int i = N; i > 0; i--)
            for (int j = M; j > 0; j--)
                if (a[i][j]) {
                    r[i][j] = r[i][j + 1] + 1, d[i][j] = d[i + 1][j] + 1;
                    // lu
                    ret += max(0, min(l[i][j], u[i][j] / 2) - 1);
                    ret += max(0, min(l[i][j], d[i][j] / 2) - 1);
                    ret += max(0, min(r[i][j], u[i][j] / 2) - 1);
                    ret += max(0, min(r[i][j], d[i][j] / 2) - 1);
                    ret += max(0, min(u[i][j], l[i][j] / 2) - 1);
                    ret += max(0, min(u[i][j], r[i][j] / 2) - 1);
                    ret += max(0, min(d[i][j], l[i][j] / 2) - 1);
                    ret += max(0, min(d[i][j], r[i][j] / 2) - 1);
                }
        cout << "Case #" << ca << ": " << ret << "
";
    }
    return 0;
}

Rabbit House (9pts, 15pts)

BFS 搜索

// RioTian 21/03/22
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 100005;
int dx[] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
int n, m, a[400][400], b[400][400];
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int T;
    cin >> T;
    for (int i = 1; i <= T; ++i) {
        cin >> n >> m;
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j) cin >> a[i][j];

        memset(b, -1, sizeof(b));
        priority_queue<pair<int, pair<int, int>>> h;
        ll cnt = 0;
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j)
                h.push(make_pair(a[i][j], make_pair(i, j)));

        while (h.size()) {
            auto t = h.top();
            h.pop();

            int x = t.second.first, y = t.second.second;
            if (b[x][y] < 0) {
                b[x][y] = t.first;
                cnt += t.first - a[x][y];
                for (int k = 0; k < 4; ++k) {
                    int nx = x + dx[k], ny = y + dy[k];
                    if (nx > 0 && nx <= n && ny > 0 && ny <= m &&
                        a[nx][ny] + 1 < b[x][y])
                        h.push(make_pair(t.first - 1, make_pair(nx, ny)));
                }
            }
        }
        cout << "Case #" << i << ": " << cnt << "
";
    }
    return 0;
}

Checksum (10pts, 17pts, 17pts)

这道题没做出来。先记录一下dalao的解法

#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <queue>

using namespace std;

const int MAXN = 100005, dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int N, M, a[505][505], b[505][505], ur[505], uc[505], fa[1005];
pair<int, int> c[505 * 505];

void init() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++) scanf("%d", &a[i][j]);
    M = 0;
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++) {
            scanf("%d", &b[i][j]);
            if (b[i][j] > 0) {
                c[++M] = make_pair(-b[i][j], (i - 1) * N + (j - 1));
            }
        }
    for (int i = 1; i <= N; i++) scanf("%d", &ur[i]);
    for (int i = 1; i <= N; i++) scanf("%d", &uc[i]);
}

int findSet(int x) {
    if (!fa[x]) return x;
    fa[x] = findSet(fa[x]);
    return fa[x];
}

void work() {
    memset(fa, 0, sizeof(fa));
    sort(c + 1, c + M + 1);
    int ret = 0;
    for (int i = 1; i <= M; i++) {
        int x = c[i].second / N + 1, y = c[i].second % N + 1 + N;
        int bx = findSet(x), by = findSet(y);
        if (bx != by) {
            if (rand() % 2) fa[bx] = by;
            else
                fa[by] = bx;
        } else
            ret -= c[i].first;
    }
    printf("%d
", ret);
}

int main() {
    // freopen("d.in", "r", stdin), freopen("d.out", "w", stdout);
    int T;
    cin >> T;
    for (int ca = 1; ca <= T; ca++) {
        printf("Case #%d: ", ca);
        init();
        work();
    }
    return 0;
}