CF1187D Subarray Sorting

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：

 #include <bits/stdc++.h>
 using namespace std;
 const int N = , INF = 0x3f3f3f3f;
 deque<int> d[N];
 int a[N], b[N], tree[N * ];

 void build(int num, int l, int r)
 {
     if (l == r) { tree[num] = a[l]; return; }
     int m = l + r >> ;
     build(num << , l, m);
     build(num <<  | , m + , r);
     tree[num] = min(tree[num << ], tree[num <<  | ]);
 }

 void update(int num, int l, int r, int x, int y)
 {
     if (l == r) { tree[num] = y; return; }
     int m = l + r >> ;
     if (x <= m) update(num << , l, m, x, y);
     else update(num <<  | , m + , r, x, y);
     tree[num] = min(tree[num << ], tree[num <<  | ]);
 }

 int query(int num, int l, int r, int x, int y)
 {
     if (x <= l && y >= r) return tree[num];
     int m = l + r >> ;
     int ans = INF;
     if (x <= m) ans = min(ans, query(num << , l, m, x, y));
     if (y >= m + ) ans = min(ans, query(num <<  | , m + , r, x, y));
     return ans;
 }

 int main()
 {
     int t, n; cin >> t;
     while (t--)
     {
         cin >> n;
         for (int i = ; i <= n; i++) d[i].clear();
         for (int i = ; i <= n; i++) { cin >> a[i]; d[a[i]].push_back(i); }
         build(, , n);
         for (int i = ; i <= n; i++) cin >> b[i];
         bool flg = true;
         int i = , j = ;
         while (i <= n && j <= n)
         {
             while (i <= n && a[i] == INF) i++;
             if (a[i] == b[j]) { d[a[i]].pop_front(); i++; j++; }
             else if (a[i] < b[j]) { flg = false; break; }
             else
             {
                 if (d[b[j]].empty()) { flg = false; break; }
                 else
                 {
                     int t = d[b[j]].front();
                     int minn = query(, , n, i, t);
                     if (minn < a[t]) { flg = false; break; }
                     else
                     {
                         update(, , n, t, INF); a[t] = INF;
                         d[b[j]].pop_front();
                         j++;
                     }
                 }
             }
         }
         cout << (flg ? "YES" : "NO") << endl;
     }
     return ;
 }