Codeforces Round #305 (Div. 二)——D

http://codeforces.com/contest/548/problem/D

D. Mike and Feet
一个长度为n的序列，在长度为x的子序列中取最小值，所有中取最大值。
for x = 1 to n

dp[i]表示以i为最小值的最长区间长度
g[i]表示区间长度为i时的最大值
sum[i]表示区间长度大于等于i的最大值

那么对于某个询问x，答案为sum[x]

单调栈搞dp[i]，O(n) 学习了姿势

#include <bits/stdc++.h>
#define clr(a,b) memset(a,b,sizeof(a))
#define clr(a,b) memset(a,b,sizeof(a))
#define Ll long long
const int maxn = 200010;
const int INF = 0x3f3f3f3f;
using namespace std;
int dp[maxn];
int l[maxn],r[maxn];
int maxl[maxn];
int x[maxn],a[maxn],val[maxn];
int g[maxn];
int sum[maxn];
int n;
void solve()
{
    stack<int> s;
    for(int i=0;i<n;++i){
        while(!s.empty()&&val[s.top()]>=val[i]){
            s.pop();
        }
        if(s.empty()) l[i]=-1;
        else l[i]=s.top();
        s.push(i);
    }

    while(!s.empty()) s.pop();
    for(int i=n-1;i>=0;--i){
        while(!s.empty()&&val[s.top()]>=val[i]){
            s.pop();
        }
        if(s.empty()) r[i]=n;
        else r[i]=s.top();
        s.push(i);
    }

    for(int i=0;i<n;++i){
//        cout<<"l[i]="<< l[i]<<" r[i]="<<r[i]<<endl;
        dp[val[i]]=max(dp[val[i]],r[i]-l[i]-1);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // ONLINE_JUDGE
    scanf("%d",&n);
    int minn=INF;
    for(int i=0;i<n;++i){
        scanf("%d",&a[i]);
        x[i]=a[i];
        minn=min(minn,a[i]);
    }
    sort(x,x+n);
    int cnt=unique(x,x+n)-x;
    for(int i=0;i<n;++i){
        val[i]=lower_bound(x,x+cnt,a[i])-x;
    }
    clr(dp,0);
    int l,r;
    solve();
//    for(int i=0;i<cnt;++i) cout<<dp[i]<<" ";cout<<endl;
    clr(g,0);
    for(int i=0;i<cnt;++i) g[dp[i]]=max(g[dp[i]],x[i]);
    sum[n]=minn;
    for(int i=n-1;i>=1;--i){
        sum[i]=max(sum[i+1],g[i]);
    }
    for(int i=1;i<=n;++i){
        printf("%d ",sum[i]);
    }
    printf("\n");
    return 0;
}