hihocoder1080 更为复杂的买卖房屋姿势

思路：

线段树区间修改，需要使用两个懒标记set和add。处理好两个标记的优先级即可（set之前的set和add是没有作用的）。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 
 const int N = ;
 
 int tree[N << ], lazy_set[N << ], lazy_add[N << ], a[N], n, q;
 
 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] = tree[num << ] + tree[num <<  | ];
 }
 
 void pushdown(int num, int cl, int cr)
 {
     if (lazy_set[num])
     {
         tree[num << ] = lazy_set[num] * cl;
         tree[num <<  | ] = lazy_set[num] * cr;
         lazy_set[num << ] = lazy_set[num];
         lazy_set[num <<  | ] = lazy_set[num];
         lazy_add[num << ] = ;
         lazy_add[num <<  | ] = ;
         lazy_set[num] = ;
     }
     if (lazy_add[num])
     {
         tree[num << ] += lazy_add[num] * cl;
         tree[num <<  | ] += lazy_add[num] * cr;
         lazy_add[num << ] += lazy_add[num];
         lazy_add[num <<  | ] += lazy_add[num];
         lazy_add[num] = ;
     }
 }
 
 void update(int num, int l, int r, int x, int y, int p, int t)
 {
     if (x <= l && y >= r)
     {
         if (t == )
         {
             tree[num] += (r - l + ) * p;
             lazy_add[num] += p;
         }
         else
         {
             tree[num] = (r - l + ) * p;
             lazy_set[num] = p;
             lazy_add[num] = ;
         }
         return;
     }
     int m = l + r >> ;
     pushdown(num, m - l + , r - m);
     if (x <= m) update(num << , l, m, x, y, p, t);
     if (y >= m + ) update(num <<  | , m + , r, x, y, p, t);
     tree[num] = 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 >> ;
     pushdown(num, m - l + , r - m);
     int ans = ;
     if (x <= m) ans += query(num << , l, m, x, y);
     if (y >= m + ) ans += query(num <<  | , m + , r, x, y);
     return ans;
 }
 
 int main()
 {
     ios::sync_with_stdio(false);
     cin >> n >> q;
     for (int i = ; i <= n + ; i++) cin >> a[i];
     build(, , n + );
     int x, y, t, p;
     for (int i = ; i <= q; i++)
     {
         cin >> t >> x >> y >> p;
         x++; y++;
         update(, , n + , x, y, p, t);
         cout << query(, , n + , , n + ) << endl;
     }
     return ;
 }