CF1187D Subarray Sorting

Posted 2022-03-11 wangyiming

思路：

线段树好题。对a数组中的每个元素从左到右依次操作，判断最终是否能够转化成b数组。在此过程中使用线段树维护区间最小值判断是否能够进行合法操作。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 300005, INF = 0x3f3f3f3f;
 4 deque<int> d[N];
 5 int a[N], b[N], tree[N * 4];
 6  
 7 void build(int num, int l, int r)
 8 
 9     if (l == r)  tree[num] = a[l]; return; 
10     int m = l + r >> 1;
11     build(num << 1, l, m);
12     build(num << 1 | 1, m + 1, r);
13     tree[num] = min(tree[num << 1], tree[num << 1 | 1]);
14 
15  
16 void update(int num, int l, int r, int x, int y)
17 
18     if (l == r)  tree[num] = y; return; 
19     int m = l + r >> 1;
20     if (x <= m) update(num << 1, l, m, x, y);
21     else update(num << 1 | 1, m + 1, r, x, y);
22     tree[num] = min(tree[num << 1], tree[num << 1 | 1]);
23 
24  
25 int query(int num, int l, int r, int x, int y)
26 
27     if (x <= l && y >= r) return tree[num];
28     int m = l + r >> 1;
29     int ans = INF;
30     if (x <= m) ans = min(ans, query(num << 1, l, m, x, y));
31     if (y >= m + 1) ans = min(ans, query(num << 1 | 1, m + 1, r, x, y));
32     return ans;
33 
34  
35 int main()
36 
37     int t, n; cin >> t;
38     while (t--)
39     
40         cin >> n;
41         for (int i = 1; i <= n; i++) d[i].clear();
42         for (int i = 1; i <= n; i++)  cin >> a[i]; d[a[i]].push_back(i); 
43         build(1, 1, n);
44         for (int i = 1; i <= n; i++) cin >> b[i];
45         bool flg = true;
46         int i = 1, j = 1;
47         while (i <= n && j <= n)
48         
49             while (i <= n && a[i] == INF) i++;
50             if (a[i] == b[j])  d[a[i]].pop_front(); i++; j++; 
51             else if (a[i] < b[j])  flg = false; break; 
52             else
53             
54                 if (d[b[j]].empty())  flg = false; break; 
55                 else
56                 
57                     int t = d[b[j]].front();
58                     int minn = query(1, 1, n, i, t);
59                     if (minn < a[t])  flg = false; break; 
60                     else
61                     
62                         update(1, 1, n, t, INF); a[t] = INF;
63                         d[b[j]].pop_front();
64                         j++;
65                     
66                 
67             
68         
69         cout << (flg ? "YES" : "NO") << endl;
70     
71     return 0;
72