Codeforces Round #504 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final)-D- Array Restoration

  我们知道不满足的肯定是两边大中间小的，这样就用RMQ查询两个相同等值的区间内部最小值即可，注意边界条件

#include<bits/stdc++.h>
#define x first
#define y second
#define ok cout << "ok" << endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const long double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const double Eps = 1e-7;
const int N = 2e5+9;

int n, q, a[N], t, pre, pos, l[N];
bool vis[N];

const int MAXN = N;
int dp[MAXN][20];
int mm[MAXN];
//初始化RMQ, b数组下标从1开始
void initRMQ(int n)
{
    mm[0] = -1;
    for(int i = 1; i <= n;i++)//
    {
        mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];//推出n在2的多少范围内
        dp[i][0] = a[i];
    }
    //RMQ操作求区间最小值
    for(int j = 1; j <= mm[n];j++)
        for(int i = 1;i + (1<<j) -1 <= n;i++)
            dp[i][j] = min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
//在区间[x, y]查询最大值
int rmq(int x,int y)
{
    int k = mm[y-x+1];
    return min(dp[x][k],dp[y-(1<<k)+1][k]);
}
int main(void) {
    while(cin >> n >> q) {
        t = 0, pos = 0;
        bool maflag = false;
        for(int i = 1; i <= n; i++) {
            scanf("%d", a + i);
            if(a[i] == q)
                maflag = true;
            if(a[i] == 0 && pos == 0)
                pos = i;
            if(a[i] == 0)
                a[i] = a[i-1];
            if(t == 0 && a[i])//防止全是0
                t = a[i];
        }
        // 都是0
        if(t == 0) {
            printf("YES\n");
            for(int i = 1; i <= n; i++) {
                printf("%d%c", q, i == n ? '\n' : ' ');
            }
            continue;
        }
        // 替代前缀0
        for(int i = 1; i <= n; i++) {
            if(a[i] == 0)
                a[i] = t;//相当于全是前导变成与第一个前导相同的
            else
                break;
        }
        // 没有出现最大值且有0
        if(!maflag && pos) {
            a[pos] = q;//最大值赋值给它
            maflag = true;
        }
        if(!maflag) {
            printf("NO\n");
            continue;
        }
        // 判断是否重复出现
        initRMQ(n);//rmq操作
        pre = a[1];
        bool flag = true;
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= n; vis[a[i]] = true, pre = a[i], l[a[i]] = i, i++) {
            if(a[i] == pre)   //a[i]用过并且令前导为a[i],a[i]的第一个编号是i
                continue;     //前面和后面一个相同
            if(vis[a[i]] && rmq(l[a[i]] + 1, i - 1) < a[i])  {
                flag = false;//如果前面已经用过a[i] 用rmq查询这两个之间是最小值如果小于a[i]就不满足
                break;
            }
        }
        if(flag) {
            printf("YES\n");
            for(int i = 1; i <= n; i++) {
                printf("%d%c", a[i], i == n ? '\n' : ' ');
            }
        }
        else
            printf("NO\n");
    }

    return 0;
}

 

来源：https://www.cnblogs.com/bluefly-hrbust/p/9497325.html