SPOJ

题意：给你2e5个矮人，编号1~N。有2e5个操作：操作1 读取x，y，交换编号为x，y的矮人。操作2 读取AB 判断编号为A，A+1····B的矮人是否连续（不必有序）。

题解：首先用pos[i]保存矮人i的位置，交换就用swap(pos[l],pos[r])来模拟。然后发现条件等价于(pos[l],pos[r])的区间满足最大值为r，最小值为l且区间内人数等于r-l+1即可。所以直接维护区间最大最小值。用change（1，p，x）来更新p处的矮人编号，并pushup。

坑：最开始ask忘写return。想用一个pair<int,int> query 一次性返回大于小于号，会tle。最坑的是忘记判端pos[l]pos[r]的大小，导致RE。

ac代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
const int maxnn = 200010;
int m, n, pos[200010];
pair<int, int> ans;
struct node {
    int l, r, maxn, minn;
}tree[800010];
inline void push_up(int x) {
    tree[x].maxn = max(tree[x << 1].maxn, tree[x << 1 | 1].maxn);
    tree[x].minn = min(tree[x << 1].minn, tree[x << 1 | 1].minn);
};


inline void build(int x, int l, int r) {
    tree[x].l = l; tree[x].r = r;
    //tree[x].maxn = l; tree[x].minn = maxnn;
    if (l == r) { tree[x].maxn = tree[x].minn = l; return; }

        int mid = (l+ r) >> 1;
        build(x << 1, l, mid);
        build(x << 1 | 1, mid + 1, r);
        push_up(x);

};
inline void change(int x, int p,int d) {
    int L=tree[x].l,R=tree[x].r;
    if (L == R) {
        tree[x].maxn = tree[x].minn = d; return;
    }

        int mid = (L + R) >> 1;
        if (p <= mid)change(x << 1,p, d);
        else change(x << 1 | 1,p, d);
        push_up(x);

}

inline int askmx(int x,int l,int r) {
    int L = tree[x].l, R = tree[x].r;
    if (L==l&&r == R)    return tree[x].maxn;
    
        int mid = (L + R) >> 1;
        if (r <= mid)return askmx(x << 1, l, r);
        else if (l > mid)return askmx(x << 1 | 1, l, r);
        else return max(askmx(x << 1, l, mid), askmx(x << 1 | 1, mid + 1, r));
}
inline int askmn(int x, int l,    int r) {
    int L = tree[x].l, R = tree[x].r;
    if (L == l&&r == R) return tree[x].minn;

        int mid = L + R >> 1;
        if (r <= mid)return askmn(x << 1, l, r);
        else if (l > mid)return askmn(x << 1 | 1, l, r);
        else return min(askmn(x << 1, l, mid), askmn(x << 1 | 1, mid + 1, r));
        
}
int main() {
    int n, q;
    cin >> n >> q;
    build(1, 1, n);
    for (int i = 1; i <= n; i++)pos[i] = i;
    
    for (int i = 1; i <= q; i++) {
        int x; int l; int r;
        scanf("%d", &x);
        scanf("%d%d", &l, &r);
        
        if (x == 1) {    
            swap(pos[l], pos[r]);
            change(1, pos[l], l);
            change(1, pos[r], r);
        }
        else {
                    
            //ans= query(1, pos[l], pos[r]);
            if (l > r)swap(l, r);
            int rr = pos[r]; 
            int ll = pos[l];
            if (ll > rr)swap(ll, rr);
            if (askmn(1, ll, rr)==l&&  askmx(1, ll, rr)==r&&rr-ll == r-l)puts("YES");
            else puts("NO");

        }
    }
}