A. Reverse

思路

签到

#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
int a[N], b[N];
int main() {
    int t;
    scanf("%d", &t);
    while (t -- ) {
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; i ++ ) {
            scanf("%d", a + i);
            b[i] = a[i];
        }
        sort(b + 1, b + 1 + n);
        int pos = 0;
        for (int i = 1; i <= n; i ++ )
            if (b[i] != a[i]) {
                pos = i;
                break;
            }
        if (pos == 0) {
            for (int i = 1; i <= n; i ++ )
                printf("%d ", a[i]);
            puts("");
            continue;
        }
        int ed;
        for (int i = 1; i <= n; i ++ ) 
            if (a[i] == b[pos]) {
                ed = i;
                break;
            }
        for (int i = pos; i <= (pos + ed) / 2; i ++ ) 
            swap(a[i], a[ed - (i - pos)]);
        for (int i = 1; i <= n; i ++ )
            printf("%d ", a[i]);
        puts("");
    }
    return 0;
}

B. Odd Swap Sort

思路

$No$ 的条件是，大奇数要跨越小奇数，大偶数要跨越小偶数。剩下的都是 $Yes$

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int sum[N];
struct Node{
    int a, id;
    int odd, even;
}num[N];
bool cmp(Node a, Node b) {
    if (a.a != b.a) return a.a < b.a;
    else return a.id < b.id;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t -- ) {
        int n;
        scanf("%d", &n);
        bool ok = 1;
        for (int i = 1; i <= n; i ++ ) {
            scanf("%d", &num[i].a);
            num[i].id = i;
            if (num[i].a & 1) {
                if (num[i - 1].odd > num[i].a) ok = 0;
                num[i].odd = max(num[i - 1].odd, num[i].a);
                num[i].even = num[i - 1].even;
            } else {
                if (num[i - 1].even > num[i].a) ok = 0;
                num[i].even = max(num[i - 1].even, num[i].a);
                num[i].odd = num[i - 1].odd;
            }
        }
        if (ok) puts("Yes");
        else puts("No");
    }
    return 0;
}

C. Inversion Graph

思路

显然这和图论并查集没有关系，并查集必然 $TLE$
那换个思路，题目考查连通块，看到样例解释的图，容易想到区间合并

#include <bits/stdc++.h>
using namespace std;
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
typedef pair<int, int> PII;
const int N = 1e5 + 10;
struct Node{
    int a, id;
    int f;
}num[N];
bool cmp(Node a, Node b) {
    return a.a < b.a;
}
int cnt;
vector<PII> seg;
void work() {
    sort(seg.begin(), seg.end());
    int ed = -2e9;
    for (auto t : seg) {
        if (ed < t.FI) cnt ++;
        ed = max(ed, t.SE);
    }
}
int main() {
    int t;
    scanf("%d", &t);
    while (t -- ) {
        int n;
        scanf("%d", &n);
        cnt = 0;
        seg.clear();
        for (int i = 1; i <= n; i ++ ) {
            scanf("%d", &num[i].a);
            num[i].id = i;
        }
        sort(num + 1, num + n + 1, cmp);
        for (int i = 1; i <= n; i ++ ) 
            num[i].f = i;
        for (int i = 1; i <= n; i ++ ) {
            seg.PB(minmax(num[i].f, num[i].id));
        }
        work();
        cout << cnt << endl;
    }
    return 0;
}

D. Big Brush

思路

$BFS$（$DFS$ 也行）
倒着跑，把一个个能删的块删掉
这题主要是细节吧，还有就是码力了，我写的比较复杂，应该有更好的，如果明天醒来还记得就接着写

#include <bits/stdc++.h>
using namespace std;
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
typedef pair<int, int> PII;
const int N = 1e3 + 10;
int g[N][N];
struct Res{
    int x, y, c;
};
vector<Res> ans;
int n, m;
int check(int x, int y) {
    VI tmp;
    if (g[x][y]) tmp.PB(g[x][y]);
    if (g[x + 1][y]) tmp.PB(g[x + 1][y]);
    if (g[x][y + 1]) tmp.PB(g[x][y + 1]);
    if (g[x + 1][y + 1]) tmp.PB(g[x + 1][y + 1]);
    if (tmp.size() == 0) return 0;
    for (int i = 1; i < tmp.size(); i ++ )
        if (tmp[i] != tmp[i - 1]) return 0;
    return tmp.size();
}
void change(int x, int y) {
    g[x][y] = g[x + 1][y] = g[x][y + 1] = g[x + 1][y + 1] = 0;
}
bool cmp(Res a, Res b) {
    return a.c > b.c;
}
void bfs(int u, int v) {
    queue<PII> q;
    q.push(MP(u, v));
    while (q.size()) {
        int x = q.front().FI, y = q.front().SE;
        q.pop();
        vector<Res> tmp;
        for (int i = -2; i < 4; i ++ ) {
            for (int j = -2; j < 4; j ++ ) {
                int a = x + i, b = y + j;
                if (a < 1 || a > n - 1 || b < 1 || b > m - 1) continue;
                int cnt = check(a, b);
                if (cnt) {
                    q.push(MP(a, b));
                    tmp.PB({a, b, cnt});
                }
            }
        }
        sort(tmp.begin(), tmp.end(), cmp);
        for (int i = 0; i < tmp.size(); i ++ ) {
            int x = tmp[i].x, y = tmp[i].y;
            int cnt = check(x, y);
            if (cnt) {
                ans.PB({x, y, max(g[x][y], max(g[x + 1][y], max(g[x][y + 1], g[x + 1][y + 1])))});
                change(x, y);
            }
        }
    }
}
int main() {
    scanf("%d%d", &n, &m);
    ans.clear();
    for (int i = 1; i <= n; i ++ ) 
        for (int j = 1; j <= m; j ++ )
            scanf("%d", &g[i][j]);
    int x = 0, y = 0;
    for (int i = 1; i < n; i ++ )
        for (int j = 1; j < m; j ++ ) {
            if (check(i, j)) {
                x = i, y = j;
                break;
            }
        }
    if (x == 0) {
        puts("-1");
        return 0;
    }
    bfs(x, y);
    bool ok = 1;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            if (g[i][j]) {
                ok = 0;
                break;
            }
    if (!ok) {
        puts("-1");
        return 0;
    }
    printf("%d\n", ans.size());
    for (int i = ans.size() - 1; i >= 0; i -- ) {
        printf("%d %d %d\n", ans[i].x, ans[i].y, ans[i].c);
    }
    return 0;
}

如果明天我醒来还记得，就接着写