Vasya And Array CF1187C（构造）

题意：

有两种操作：

① $1$ $l$ $r$ 表示区间$(l,r)$中不递减。

② $0$ $l$ $r$ 表示区间$(l,r)$中存在递减。

要求构造这个序列。

思路:

对于$1,l,r$中的每一个数等于前一个数，其余的数$a[i]=a[i-1]-1$。

对于$0,l,r$判断$a[l]$和$a[r]$的关系，若相等则不存在可构造序列。

代码：

//#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define pb push_back

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;

int n, m;
int a[maxn];
struct node
{
    int op, l, r;
}p[maxn];

int vis[maxn];

int main()
{
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= m; ++i)
    {
        scanf("%d %d %d", &p[i].op, &p[i].l, &p[i].r);
        if (p[i].op == 1)
            for (int j = p[i].l + 1; j <= p[i].r; ++j)    vis[j] = 1;
    }
    a[1] = n;
    for (int i = 2; i <= n; ++i)
    {
        if (vis[i] == 1)    a[i] = a[i - 1];
        else a[i] = a[i - 1] - 1;
    }
    for (int i = 1; i <= m; ++i)
    {
        if (p[i].op)    continue;
        if (a[p[i].l] == a[p[i].r]) {
            cout << "NO" << endl;
            return 0;
        }
    }
    cout << "YES" << endl;
    for (int i = 1; i <= n; ++i)
    {
        if (i != 1)    cout << " ";
        cout << a[i];
    }
}