专题测试二 树形结构 A - Circular RMQ

1. 题目
   

   You are given circular array a₀, a₁, ..., a_n - 1. There are two types of operations with it:

   • inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
   • rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).

   Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.

   Write program to process given sequence of operations.

   Input

   The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a₀, a₁, ..., a_n - 1 ( - 10⁶ ≤ a_i ≤ 10⁶), a_i are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 10⁶ ≤ v ≤ 10⁶) — inc operation.

   Output

   For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

   Sample
   

   Input	Output
   4 1 2 3 4 4 3 0 3 0 -1 0 1 2 1	1 0 0

2. 思路
   线段树，但是数组是环形的，而且和不少人习惯不一样，数组从0下标开始
   在线段树模板上增加判断，如果l>r的话，就搜l~n-1和0~r的和，否则就正常处理
   要方便的话直接把输入的操作区间全部+1就是从1下标开始了
3. 代码
   

   #include<cstdio>
   #include<iostream>
   #include<string>
   #include<cstring>
   #include<algorithm>
   using namespace std;
   
   #define ll long long
   
   ll d[8000000],b[8000000],a[2000005];//区间值，懒惰标记，原始值
   ll n,q;
   
   void build(ll s, ll t, ll p) {
   
   	if (s == t) {
   	    d[p] = a[s];
   	    return;
   	}
   
   	ll m = s + ((t - s) >> 1);
   	build(s, m, p << 1), build(m + 1, t, p << 1 | 1);
   	d[p] =min(d[p << 1], d[(p << 1) + 1]);
   }
   
   void update(ll l, ll r, ll c, ll s, ll t, ll p) {
   
   	if (l <= s && t <= r) {
   		d[p] += c, b[p] += c;
   	    return;
   	}
   
   	ll m = s + ((t - s) >> 1);
   	if (b[p] && s != t) {
   	    d[p << 1] += b[p], d[p << 1 | 1] += b[p];
   		b[p << 1] += b[p], b[p << 1 | 1] += b[p];
   		b[p] = 0;
   	}
   	if (l <= m) update(l, r, c, s, m, p << 1);
   	if (r > m) update(l, r, c, m + 1, t, p << 1 | 1);
   	d[p] = min(d[p << 1], d[p << 1 | 1]);
   }
   
   ll query(ll l, ll r, ll s, ll t, ll p) {
   
   	if (l <= s && t <= r) return d[p];
   
   	ll m = s + ((t - s) >> 1);
   	if (b[p]) {
   	    d[p << 1] += b[p], d[p << 1 | 1] += b[p];
           b[p << 1] += b[p], b[p << 1 | 1] += b[p];
   		b[p] = 0;
   	}
   	ll ans = 10e8;
   	if (l <= m) ans = min(ans,query(l, r, s, m, p << 1));
   	if (r > m) ans = min(ans,query(l, r, m + 1, t, p << 1 | 1));
   	return ans;
   }
   
   int main()
   {
   	cin >> n;
   	for(int i=1;i<=n;i++){
   		cin >> a[i];
   	}
   	build(1,n,1);
   	cin >> q;
   	ll x,y,z;
       while(q--)
       {
           cin >> x >> y;
           x++;y++;//题目数组从0开始
           if(getchar() != '\n')//没换行是三个数
           {
               cin >> z;
               if(x <= y) update(x, y, z, 1, n, 1);
               else
               {
                   update(x, n, z, 1, n, 1);
                   update(1, y, z, 1, n, 1);
               }
           }
           else
           {
               if(x <= y)
   			cout << query(x, y, 1, n, 1) << endl;
               else
   			cout << min(query(x, n, 1, n, 1), query(1, y, 1, n, 1)) << endl;
           }
       }
   }