CF 900.C Remove Extra One 单调栈+BIT

题意:[1..n]的排列a. 定义a[i]合法:当所有j (1<=j<i)都满足a[j]<a[i].
n<=1e5.删除一个数之后 要使得排列a的合法个数最多 输出要删除的哪一个数.若有多解,输出最小的.


定义f[i]为删除第i个数 能让答案的贡献增大多少,c[i]为第i个元素左边比它大的个数
先用单调栈处理出每个元素左边第一个比它的大元素.

当c[i]等于1 删除第i个元素左边第一个比它大的元素j 答案才有可能增加 让f[j]++,求出最大的f[j]即可.

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N=2e5+5;int n,a[N],b[N],L[N],c[N],p[N],f[N];stack<int> s;int lowbit(int x){return x&-x;}void update(int x,int val){for(int i=x;i<N;i+=lowbit(i))b[i]+=val;}int sum(int x){int res=0;for(int i=x;i>0;i-=lowbit(i))res+=b[i];return res;}int main(){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);while(!s.empty()&&a[s.top()]<a[i])s.pop();//R[s.top]=iL[i]=s.empty()?0:s.top();s.push(i);p[a[i]]=i;}for(int i=n;i>=1;i--){update(p[i],1);c[i]=sum(p[i]-1);//cout<<c[i]<<' '<<p[i]<<endl;}for(int i=1;i<=n;i++){if(c[i]>1)continue;if(c[i]==0)f[i]--;if(c[i]==1)f[a[L[p[i]]]]++;}int res=1,mx=-2;for(int i=1;i<=n;i++){if(mx<f[i])mx=f[i],res=i;}cout<<res<<endl;return 0;}

更简单的做法 代码来自Benq

#include <bits/stdc++.h>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/assoc_container.hpp>using namespace std;using namespace __gnu_pbds; typedef long long ll;typedef vector<int> vi;typedef pair<int, int> pii;template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;#define FOR(i, a, b) for (int i=a; i<(b); i++)#define F0R(i, a) for (int i=0; i<(a); i++)#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)#define sz(x) (int)(x).size()#define mp make_pair#define pb push_back#define f first#define s second#define lb lower_bound#define ub upper_bound#define all(x) x.begin(), x.end()const int MOD = 1000000007;int n, p[100000], ok[100000], co[100000];set<pii> a;pii bes = {-MOD,-MOD};int main() {ios_base::sync_with_stdio(0);cin.tie(0);cin >> n;F0R(i,n) cin >> p[i];F0R(i,n) {    if (!sz(a) || a.rbegin()->f < p[i]) {        ok[i] = -1;    } else if (sz(a) < 2 || next(a.rbegin())->f < p[i]) {        co[a.rbegin()->s] ++;    }    a.insert({p[i],i});}F0R(i,n) bes = max(bes,mp(co[i]+ok[i],-p[i]));cout << -bes.s;}