set+线段树 Codeforces Round #305 (Div. 2) D. Mike and Feet

题目传送门

 /*
     题意：对于长度为x的子序列，每个序列存放为最小值，输出长度为x的子序列的最大值
     set+线段树：线段树每个结点存放长度为rt的最大值，更新：先升序排序，逐个添加到set中
                 查找左右相邻的位置，更新长度为r - l - 1的最大值，感觉线段树结构体封装不错！
     详细解释：http://blog.csdn.net/u010660276/article/details/46045777
     其实还有其他解法，先掌握这种:)
 */
 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <iostream>
 #include <cstring>
 #include <set>
 using namespace std;

 typedef long long ll;
 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1

 const int MAXN = 2e5 + ;
 const int INF = 0x3f3f3f3f;

 struct A
 {
     int v, id;
     bool operator < (const A &a) const
     {
         return v < a.v;
     }
 }a[MAXN];
 struct SegmentTree
 {
     int add[MAXN << ];
     int mx[MAXN << ];

     void push_down(int rt)
     {
         if (add[rt])
         {
             add[rt<<] = add[rt<<|] = add[rt];
             mx[rt<<] = mx[rt<<|] = add[rt];
             add[rt] = ;
         }
     }

     void build(int l, int r, int rt)
     {
         add[rt] = mx[rt] = ;
         if (l == r)    return ;
         int mid = (l + r) >> ;
         build (lson);    build (rson);
     }

     void updata(int ql, int qr, int c, int l, int r, int rt)
     {
         if (ql <= l && r <= qr)    {mx[rt] = c; add[rt] = c;    return ;}
         push_down (rt);
         int mid = (l + r) >> ;
         if (ql <= mid)    updata (ql, qr, c, lson);
         if (qr > mid)    updata (ql, qr, c, rson);
     }

     int query(int x, int l, int r, int rt)
     {
         if (l == r)    return mx[rt];
         push_down (rt);
         int mid = (l + r) >> ;
         if (x <= mid)    return query (x, lson);
         return query (x, rson);
     }
 }tree;

 int ans[MAXN];

 int main(void)        //Codeforces Round #305 (Div. 2) D. Mike and Feet
 {
     int n;
     while (scanf ("%d", &n) == )
     {
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i].v);    a[i].id = i;
         }
         sort (a+, a++n);

         tree.build (, n, );
         set<int> S;    S.insert ();    S.insert (n + );
         set<int>::iterator it;
         for (int i=; i<=n; ++i)
         {
             int l, r;
             it = S.lower_bound (a[i].id);
             r = *it; it--; l = *it;
             if (r - l -  >= )    tree.updata (, r-l-, a[i].v, , n, );
             S.insert (a[i].id);
         }

         for (int i=; i<=n; ++i)
             ans[i] = tree.query (i, , n, );
         for (int i=; i<=n; ++i)
             printf ("%d%c", ans[i], (i==n) ? '\n' : ' ');
     }

     return ;
 }

 /*
 10
 1 2 3 4 5 4 3 2 1 6
 */

 

 /*
     jinye的解法：原理一样，找到最小值是a[i]的最长区间，用l[],r[]记录左右端点的位置
                 比线段树快多了:)
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 const int MAXN = 2e5 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN], l[MAXN], r[MAXN], ans[MAXN];

 int main(void)        //Codeforces Round #305 (Div. 2) D. Mike and Feet
 {
     int n;
     while (scanf ("%d", &n) == )
     {
         memset (ans, , sizeof (ans));
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i]);
             l[i] = r[i] = i;
         }

         for (int i=; i<=n; ++i)
         {
             while (l[i] >  && a[i] <= a[l[i]-])    l[i] = l[l[i]-];
         }
         for (int i=n; i>=; --i)        //倒过来回超时
         {
             while (r[i] < n && a[i] <= a[r[i]+])    r[i] = r[r[i]+];
         }

         for (int i=; i<=n; ++i)
         {
             int x = r[i] - l[i] + ;
             ans[x] = max (ans[x], a[i]);
         }

         for (int i=n-; i>=; --i)        //可能范围扩展的太厉害了
         {
             ans[i] = max (ans[i], ans[i+]);
         }

         for (int i=; i<=n; ++i)
             printf ("%d%c", ans[i], (i==n) ? '\n' : ' ');
     }

     return ;
 }

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