Sequence（线段树+二分）

传送门

第二次做这种题目，把人做傻了，没想到是二分，只是隐隐约约感觉到了二分的影子。于是写了两个查找的函数，但是发现查找的函数无法解决求左边最大值和右边最小值的问题。写了一大堆烂代码。
正解只需要在查询的时候不断二分就行了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 1e5 + 10;
struct Node{
    int l, r, minn, maxn;
}node[N << 2];
int n, m, a[N];
void pushup(int root){
    node[root].minn = min(node[root << 1].minn, node[root << 1 | 1].minn);
    node[root].maxn = max(node[root << 1].maxn, node[root << 1 | 1].maxn);
}
void build(int root, int l, int r){
    node[root].l = l;
    node[root].r = r;
    if(l == r){
        node[root].minn = a[l];
        node[root].maxn = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(root << 1, l, mid);
    build(root << 1 | 1, mid + 1, r);
    pushup(root);
}
void update(int root, int pos, int value){
    if(node[root].l == node[root].r){
        node[root].minn = value;
        node[root].maxn = value;
        return;
    }
    int mid = (node[root].l + node[root].r) >> 1;
    if(pos <= mid)
        update(root << 1, pos, value);
    else
        update(root << 1 | 1, pos, value);
    pushup(root);
}
/*ll queryleft(int root, int l, int r, int value){
    if(node[root].l == node[root].r){
        return node[root].l;
    }
    int mid = (node[root].l + node[root].r) >> 1;
    if(node[root].l >= l && node[root].r <= r){
        if(node[root].minn > value){
            return node[root].l;
        }else if(node[root].maxn < value){
            return node[root].l - 1;
        }else if(node[root].minn < value && node[root].maxn >= value){
            if(node[root << 1].minn < value)
            	return queryleft(root << 1, l, r, value);
            else if(node[root << 1 | 1].minn < value)
            	return queryleft(root << 1 | 1, l, r, value);
        }
    }
    ll ans = -1;
    if(l <= mid)
        ans = max(ans, queryleft(root << 1, l, r, value));
    //cout << "ansleft1 = " << ans << endl;
    if(r > mid)
        ans = max(ans, queryleft(root << 1 | 1, l, r, value));
    //cout << "ansright1 = " << ans << endl;
	return ans;
}
int queryright(int root, int l, int r, int value){
	if(node[root].l == node[root].r){
        return node[root].l;
    }
    int mid = (node[root].l + node[root].r) >> 1;
    if(node[root].l >= l && node[root].r <= r){
        if(node[root].minn > value){
            //return node[root].r;
        	return queryright(root << 1 | 1, l, r, value);
		}else if(node[root].maxn < value){
            return node[root].l - 1;
        }else if(node[root].minn < value && node[root].maxn >= value){
        	if(node[root << 1].minn < value)
        		return queryright(root << 1, l, r, value);
        	else if(node[root << 1 | 1].minn < value)
        		return queryright(root << 1 | 1, l, r, value);
        }
    }
    int ans = 0x3f3f3f3f;
    if(l <= mid)
        ans = min(ans, queryright(root << 1, l, r, value));
    cout << "ansleft2 = " << ans << endl;
    if(r > mid)
        ans = min(ans, queryright(root << 1 | 1, l, r, value));
    cout << "ansright2 = " << ans << endl;
    return ans;
}*/
ll query(int root, int l, int r){
	if(node[root].l >= l && node[root].r <= r){
		return node[root].minn;
	} 
	int mid = (node[root].l + node[root].r) >> 1;
	ll ans = 1e9 + 10;
	if(l <= mid)
		ans = min(ans, query(root << 1, l, r));
	if(r > mid)
		ans = min(ans, query(root << 1 | 1, l, r));
	return ans;
}
int main(){
    cin >> n >> m;
    for(int i = 1; i <= n; i ++)
        cin >> a[i];
    build(1, 1, n);
    for(int i = 1; i <= m; i ++){
        int op;
        cin >> op;
        if(op == 1){
            int x, y;
            cin >> x >> y;
            a[x] = y;
            update(1, x, y);
        }else{
            int pos;
            cin >> pos;
            int value = a[pos];
            int l = 1, r = pos;
            while(l < r){
            	int mid = (l + r) >> 1;
            	if(query(1, mid, pos) >= a[pos])
            		r = mid;
            	else
            		l = mid + 1;
			}
			int ansleft = pos - l + 1;
			l = pos, r = n + 1;
			while(l < r){
				int mid = (l + r) >> 1;
				if(query(1, pos, mid) < a[pos])
					r = mid;
				else
					l = mid + 1;
			}
			int ansright = l - pos;
            cout<< (ll)ansleft * ansright << endl;
        }
    }
    return 0;
}