codeforces 631C. Report

题目链接

按题目给出的r， 维护一个递减的数列，然后在末尾补一个0。 比如样例给出的

4 2
1 2 4 3
2 3
1 2

递减的数列就是3 2 0， 操作的时候， 先变[3, 2), 然后变[2, 0)， 具体的过程看代码。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 2e5+5;
int st[200005], a[maxn], b[maxn], c[maxn];
int main()
{
    int n, m, sign, pos;
    cin>>n>>m;
    for(int i = 1; i<=n; i++)
        scanf("%d", &a[i]);
    int r = 0;
    for(int i = 0; i<m; i++) {
        scanf("%d%d", &sign, &pos);
        while(r>0&&pos>=st[r-1])
            r--;
        st[r] = pos, b[r] = sign;
        r++;
    }
    st[r++] = 0;
    int l = 1, rr = st[0];
    for(int i = 1; i<=n; i++)
        c[i] = a[i];
    sort(c+1, c+1+rr);
    for(int i = 1; i<r; i++) {
        for(int j = st[i-1]; j>st[i]; j--) {
            a[j] = (b[i-1] == 2)?c[l++]:c[rr--];
        }
    }
    for(int i = 1; i<=n; i++)
        printf("%d ", a[i]);
    return 0;
}