最后一场多校打得一般般。
题目链接：http://acm.hdu.edu.cn/contests/contest_show.php?cid=857
C：
E：
I：
BFS水题。
1 /* Codeforces Contest 2019_mutc_10
2 * Problem I
3 * Au: SJoshua
4 */
5 #include <queue>
6 #include <cstdio>
7 #include <vector>
8 #include <string>
9 #include <iostream>
10
11 using namespace std;
12
13 bool board[2002][2002];
14 int n, m, q;
15
16 const int movement[4][2] = {
17 {0, 1}, {0, -1}, {1, 0}, {-1, 0}
18 };
19
20 struct pos {
21 int x, y;
22 };
23
24 int solve(int x, int y) {
25 int ans = 0;
26 if (board[x][y]) {
27 queue <pos> q;
28 q.push({x, y});
29 while (!q.empty()) {
30 auto t = q.front();
31 q.pop();
32 if (!board[t.x][t.y]) {
33 continue;
34 }
35 ans++;
36 board[t.x][t.y] = 0;
37 for (int i = 0; i < 4; i++) {
38 int nx = t.x + movement[i][0], ny = t.y + movement[i][1];
39 if (1 <= nx && nx <= n && 1 <= ny && ny <= m && board[nx][ny]) {
40 if ((!board[nx + 1][ny] || !board[nx - 1][ny]) && (!board[nx][ny - 1] || !board[nx][ny + 1])) {
41 q.push({nx, ny});
42 }
43 }
44 }
45 }
46 }
47 return ans;
48 }
49
50 int main(void) {
51 int T;
52 scanf("%d", &T);
53 while (T--) {
54 scanf("%d %d %d", &n, &m, &q);
55 for (int i = 0; i <= n + 1; i++) {
56 for (int j = 0; j <= m + 1; j++) {
57 board[i][j] = true;
58 }
59 }
60 while (q--) {
61 int x, y;
62 scanf("%d %d", &x, &y);
63 printf("%d\n", solve(x, y));
64 }
65 }
66 return 0;
67 }
K：
做法略像之前牛客多校第四场C。对于每个a[i]，计算出在满足区间内元素unique的前提下，往左往右能到的最大区间。
考虑启发式分治，对于当前区间，找到最大值所在位置并记为mid并看mid是更靠左还是靠右。若更靠左，则