2019 HDOJ Multi-University Training Contest Stage 10(杭电多校)

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

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

原文地址：https://www.cnblogs.com/JHSeng/p/11409524.html