PAT (Top Level) Practise 1017 The Best Peak Shape (35)

1017. The Best Peak Shape (35)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

In many research areas, one important target of analyzing data is to find the best "peak shape" out of a huge amount of raw data full of noises. A "peak shape" of length L is an ordered sequence of L numbers { D₁, ..., D_L } satisfying that there exists an index i (1 < i < L) such that D₁ < ... < D_i-1 < D_i > D_i+1 > ... > D_L.

Now given N input numbers ordered by their indices, you may remove some of them to keep the rest of the numbers in a peak shape. The best peak shape is the longest sub-sequence that forms a peak shape. If there is a tie, then the most symmetric (meaning that the difference of the lengths of the increasing and the decreasing sub-sequences is minimized) one will be chosen.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (3 <= N <= 10⁴). Then N integers are given in the next line, separated by spaces. All the integers are in [-10000, 10000].

Output Specification:

For each case, print in a line the length of the best peak shape, the index (starts from 1) and the value of the peak number. If the solution does not exist, simply print "No peak shape" in a line. The judge's input guarantees the uniqueness of the output.

Sample Input 1:

201 3 0 8 5 -2 29 20 20 4 10 4 7 25 18 6 17 16 2 -1

Sample Output 1:

10 14 25

Sample Input 2:

5-1 3 8 10 20

Sample Output 2:

No peak shape

本场顶级压轴题,然后却没什么难度。

给出n个数字，要求找出一个序列满足先上升再下降（严格的），求最长的长度，并且同长度要求左右长度差最小。

其实就是求最长上升子序列，对于一个位置i，我们可以从前往后求出以i为结尾的最长上升子序列的长度，同样可以从后往前再求一遍。

这样两边加起来减一就是这个点的答案，然后在比较一下那个比较优就行了。求最长上升什么的可以用树状数组或者二分。

#include<map> #include<set>#include<ctime>  #include<cmath>      #include<queue>   #include<string>  #include<vector>  #include<cstdio>      #include<cstring>    #include<iostream>  #include<algorithm>      #include<functional>  using namespace std;#define ms(x,y) memset(x,y,sizeof(x))      #define rep(i,j,k) for(int i=j;i<=k;i++)      #define per(i,j,k) for(int i=j;i>=k;i--)      #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])      #define inone(x) scanf("%d",&x)      #define intwo(x,y) scanf("%d%d",&x,&y)      #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)    #define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)   #define lson x<<1,l,mid  #define rson x<<1|1,mid+1,r  #define mp(i,j) make_pair(i,j)  #define ft first  #define sd second  typedef long long LL;typedef pair<int, int> pii;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 3e5 + 10;const int M = 1e4 + 1;const double eps = 1e-10;int n, a[N], l[N], r[N], f[N];int get(int x) { int res = 0; for (int i = x; i; i -= low(i)) res = max(res, f[i]); return res; }void add(int x, int y) { for (int i = x; i < N; i += low(i)) f[i] = max(f[i], y); }int main(){  inone(n);  rep(i, 1, n) inone(a[i]), a[i] += M;  rep(i, 1, n)  {    l[i] = get(a[i] - 1) + 1;    add(a[i], l[i]);  }  ms(f, 0);  per(i, n, 1)  {    r[i] = get(a[i] - 1) + 1;    add(a[i], r[i]);  }  int ans = 0, index = 0, dis = 0;  rep(i, 1, n)  {    if (l[i] < 2 || r[i] < 2) continue;    if (ans < l[i] + r[i] - 1)    {      ans = l[i] + r[i] - 1;      index = i; dis = abs(l[i] - r[i]);    }    else if (ans == l[i] + r[i] - 1 && dis > abs(l[i] - r[i]))    {      index = i; dis = abs(l[i] - r[i]);    }  }  if (!ans) puts("No peak shape");  else printf("%d %d %d\n", ans, index, a[index] - M);  return 0;}