2019年杭电多校第二场 1002题Beauty Of Unimodal Sequence(LIS+单调栈)

题目链接

传送门

思路

首先我们对\(a\)正反各跑一边\(LIS\),记录每个位置在前一半的\(LIS\)中应该放的位置\(ans1[i]\),后一半的位置\(ans2[i]\)

对于字典序最小的方案,我们找到第一个峰值,然后往前遍历。在\(i\)这个位置,如果它在\(LIS\)中放的位置是\(pos\),那么我们先看当前放在\(pos+1\)的值是否比它大,大的话就说明这个位置一定比前面放过在\(pos\)这个位置的更优(因为字典序更小,且\([1,i]\)一定可以放满\([1,pos-1]\)),然后我们就用单调栈把当前放的小于等于\(pos\)的值全部\(pop\)掉。对于后一半的我们就直接对于每个可以放的位置反复覆盖即可。

对于字典序最大的我们则就是前一半反复覆盖,后一半用单调栈维护,与字典序刚好相反。

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 998244353;
const int maxn = 3e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n;
stack<int> st;
vector<int> vec, v;
int pos[maxn];
int a[maxn], dp[maxn], ans1[maxn], ans2[maxn];

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    while(~scanf("%d", &n)) {
        memset(dp, inf, sizeof(dp));
        dp[0] = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            ans1[i] = ans2[i] = 0;
            pos[i] = inf;
            ans1[i] = lower_bound(dp + 1, dp + n + 1, a[i]) - dp;
            dp[ans1[i]] = a[i];
        }
        memset(dp, inf, sizeof(dp));
        dp[0] = 0;
        for(int i = n; i >= 1; --i) {
            ans2[i] = lower_bound(dp + 1, dp + n + 1, a[i]) - dp;
            dp[ans2[i]] = a[i];
        }
        while(!st.empty()) st.pop();
        int idx = 1, mx = ans1[1] + ans2[1];
        for(int i = 2; i <= n; ++i) {
            if(ans1[i] + ans2[i] > mx) {
                idx = i;
                mx = ans1[i] + ans2[i];
            }
        }
        pos[ans1[idx]] = a[idx];
        for(int i = idx - 1; i >= 1; --i) {
            if(ans1[i] >= ans1[idx]) continue;
            if(pos[ans1[i]+1] <= a[i]) continue;
            while(!st.empty() && ans1[i] >= ans1[st.top()]) {
                pos[ans1[st.top()]] = inf;
                st.pop();
            }
            st.push(i);
            pos[ans1[i]] = a[i];
        }
        vec.clear();
        while(!st.empty()) {
            vec.push_back(st.top());
            st.pop();
        }
        vec.push_back(idx);
        int las = idx;
        for(int i = idx + 1; i <= n; ++i) {
            if(ans2[i] == ans2[las] - 1 && a[i] < a[las]) {
                vec.push_back(i);
                las = i;
            }
        }
        int flag = 0;
        for(auto x:vec) {
            if(flag) printf(" ");
            flag = 1;
            printf("%d", x);
        }
        printf("\n");

        vec.clear();
        idx = 1, mx = ans1[1] + ans2[1];
        for(int i = 2; i <= n; ++i) {
            if(ans1[i] + ans2[i] >= mx) {
                idx = i;
                mx = ans1[i] + ans2[i];
            }
        }
        las = idx;
        vec.push_back(idx);
        for(int i = idx - 1; i >= 1; --i) {
            if(ans1[i] == ans1[las] - 1 && a[i] < a[las]) {
                vec.push_back(i);
                las = i;
            }
        }
        reverse(vec.begin(), vec.end());
        for(int i = 1; i<= n; ++i) pos[i] = inf;
        pos[ans2[idx]] = a[idx];
        while(!st.empty()) st.pop();
        for(int i = idx + 1; i <= n; ++i) {
            if(ans2[i] > ans2[idx]) continue;
            if(a[i] >= pos[ans2[i]+1]) continue;
            while(!st.empty() && ans2[i] >= ans2[st.top()]) {
                pos[ans2[st.top()]] = inf;
                st.pop();
            }
            st.push(i);
            pos[ans2[st.top()]] = a[i];
        }
        v.clear();
        while(!st.empty()) {
            v.push_back(st.top());
            st.pop();
        }
        reverse(v.begin(), v.end());
        flag = 0;
        for(auto x:vec) {
            if(flag) printf(" ");
            flag = 1;
            printf("%d", x);
        }
        for(auto x:v) {
            if(flag) printf(" ");
            flag = 1;
            printf("%d", x);
        }
        printf("\n");
    }
    return 0;
}
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posted @ 2019-07-24 18:34  Dillonh  阅读(571)  评论(0编辑  收藏  举报