poj1363 Rails Central Europe 1997

P.S.:

输出换行

三个方法

1.直接按照要求做

根据给的数，需要push,pop哪些数据，具有唯一性

数最多进栈一次，出栈一次

O(n)

 Source Code
 Problem:         User: congmingyige
 Memory: 728K        Time: 63MS
 Language: G++        Result: Accepted

     Source Code

     #include <cstdio>
     #include <cstdlib>
     #include <cmath>
     #include <cstring>
     #include <string>
     #include <algorithm>
     #include <iostream>
     using namespace std;
     #define ll long long

     const double eps=1e-;
     const ll inf=1e9;
     const ll mod=1e9+;
     const int maxn=1e5+;

     int a[maxn];
     int st[maxn];

     int main()
     {
         int n,i,j,g;
         bool line=;
         while ()
         {
             scanf("%d",&n);
             if (n==)
                 break;

             if (!line)
                 line=;
             else
                 printf("\n");

             g=;
             while ()
             {
                 scanf("%d",&a[]);
                 if (a[]==)
                     break;
                 for (i=;i<=n;i++)
                     scanf("%d",&a[i]);

                 j=;
                 for (i=;i<=n;i++)
                 {
                     while (j<a[i])
                         st[++g]=++j;
                     if (st[g]==a[i])
                         g--;
                     else
                         break;
                 }

                 if (i==n+)
                     printf("Yes\n");
                 else
                     printf("No\n");
             }
         }
         return ;
     }
     /*
     6
     1 3 2 4 6 5
     1 4 3 5 2 6
     1 3 6 5 2 4
     5 4 3 2 1 6
     5 3 2 1 6 4
     1 3 5 4 6 2
     1 3 6 2 5 4

     4
     1 2 3 4
     1 2 4 3
     1 3 2 4
     1 3 4 2
     1 4 2 3
     1 4 3 2
     2 1 3 4
     2 1 4 3
     2 3 1 4
     2 3 4 1
     2 4 1 3
     2 4 3 1
     3 1 2 4
     3 1 4 2
     3 2 1 4
     3 2 4 1
     3 4 1 2
     3 4 2 1
     4 1 2 3
     4 1 3 2
     4 2 1 3
     4 2 3 1
     4 3 1 2
     4 3 2 1

     2 3 1 4
     */

2.

一个序列，数x在第y个位置，第y个位置之后的小于x的数，必须是越往右越小。

即当前的最大数假设为z，则小于z的数必须是z-1,z-2,..,1。

 

时间上为O(n)，但常数较大

 

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define ll long long

 const double eps=1e-;
 const ll inf=1e9;
 const ll mod=1e9+;
 const int maxn=1e5+;

 int a[maxn];
 bool vis[maxn];

 int main()
 {
     int n,i,j;
     bool line=;
     while ()
     {
         scanf("%d",&n);
         if (n==)
             break;

         if (!line)
             line=;
         else
             printf("\n");

         while ()
         {
             scanf("%d",&a[]);
             if (a[]==)
                 break;
             for (i=;i<=n;i++)
                 scanf("%d",&a[i]);

             memset(vis,,sizeof(vis));
             j=;
             for (i=;i<=n;i++)
             {
                 if (j>a[i])
                     break;
                 else if (j<=a[i])
                     j=a[i]-;

                 vis[a[i]]=;
                 while (vis[j])
                     j--;
             }

             if (i==n+)
                 printf("Yes\n");
             else
                 printf("No\n");
         }
     }
     return ;
 }
 /*
 6
 1 3 2 4 6 5
 1 4 3 5 2 6
 1 3 6 5 2 4
 5 4 3 2 1 6
 5 3 2 1 6 4
 1 3 5 4 6 2
 1 3 6 2 5 4

 8
 1 3 6 8 7 5 4 2

 4
 1 2 3 4
 1 2 4 3
 1 3 2 4
 1 3 4 2
 1 4 2 3
 1 4 3 2
 2 1 3 4
 2 1 4 3
 2 3 1 4
 2 3 4 1
 2 4 1 3
 2 4 3 1
 3 1 2 4
 3 1 4 2
 3 2 1 4
 3 2 4 1
 3 4 1 2
 3 4 2 1
 4 1 2 3
 4 1 3 2
 4 2 1 3
 4 2 3 1
 4 3 1 2
 4 3 2 1

 2 3 1 4
 */

previous数组

一个数跳出，使用previous数组，这个数不再被使用

O(n)

1 3 2 6 5 4 9 8 7

则

previous[4] 0

previous[6] 6->5->4->0 0

previous[7] 7->6->0

极限数据

1 2 3 4 5 6 7 8 9

按照上述方法O(n^2)

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define ll long long

 const double eps=1e-;
 const ll inf=1e9;
 const ll mod=1e9+;
 const int maxn=1e5+;

 int a[maxn],pre[maxn];
 bool vis[maxn];

 int main()
 {
     int n,i,j,k;
     bool line=;
     while ()
     {
         scanf("%d",&n);
         if (n==)
             break;

         if (!line)
             line=;
         else
             printf("\n");

         while ()
         {
             scanf("%d",&a[]);
             if (a[]==)
                 break;
             for (i=;i<=n;i++)
                 scanf("%d",&a[i]);

             memset(vis,,sizeof(vis));
             j=;
             for (i=;i<=n;i++)
             {
                 if (j>a[i])
                     break;
                 else if (j<=a[i])
                     j=a[i]-,k=j;

                 vis[a[i]]=;
                 while (vis[j])
                 {
                     if (pre[j])
                         j=pre[j];
                     else
                         j--;
                 }
                 if (j!=k)
                     pre[k]=j;
             }

             if (i==n+)
                 printf("Yes\n");
             else
                 printf("No\n");
         }
     }
     return ;
 }
 /*
 6
 1 3 2 4 6 5
 1 4 3 5 2 6
 1 3 6 5 2 4
 5 4 3 2 1 6
 5 3 2 1 6 4
 1 3 5 4 6 2
 1 3 6 2 5 4

 8
 1 3 6 8 7 5 4 2

 4
 1 2 3 4
 1 2 4 3
 1 3 2 4
 1 3 4 2
 1 4 2 3
 1 4 3 2
 2 1 3 4
 2 1 4 3
 2 3 1 4
 2 3 4 1
 2 4 1 3
 2 4 3 1
 3 1 2 4
 3 1 4 2
 3 2 1 4
 3 2 4 1
 3 4 1 2
 3 4 2 1
 4 1 2 3
 4 1 3 2
 4 2 1 3
 4 2 3 1
 4 3 1 2
 4 3 2 1

 2 3 1 4
 */