标签：des style c class blog code
| Longest Ordered Subsequence Extention |
| Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB |
| Total submit users: 778, Accepted users: 555 |
| Problem 10027 : No special judgement |
| Problem description |
|
A numeric sequence
of ai is ordered if a1 <
a2 < ... < aN. Let the
subsequence of the given numeric sequence (a1,
a2, ..., aN) be any sequence
(ai1, ai2,
..., aiK), where 1 <=
i1 < i2 < ... <
iK <= N. For example, sequence (1, 7, 3,
5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many
others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5,
8). |
| Input |
|
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 100000 each, separated by spaces. 1 <= N <= 50000 |
| Output |
|
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence. |
| Sample Input |
7 1 7 3 5 9 4 8 |
| Sample Output |
4 |
起初，我愚蠢的以为这道题和10001一样，仅仅把数组开大，然后就WA了。
很感谢yiyi学长教我写这道题。n=50000，要优化为一个O(n*logn)的DP，用到二分查找。
二分法是中学解方程的时候常用的方法，也是ACM中查找某个满足条件的值的方法，当数据量很大适宜采用该方法。注意的是，采用二分法查找时，数据需是排好序的。
拿样例来说，b数组中元素的变化：
1 1 7 1 3 1 3 5 1 3 5 9 1 3 4 9 1 3 4 8
因此 k = 4 。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<algorithm> 6 #include<cmath> 7 8 using namespace std; 9 10 int Check(int *p); 11 12 int N; 13 int i, j, k; 14 int a[50005]; 15 int b[50005]; 16 17 int main() 18 { 19 while(scanf("%d",&N)!=EOF) { 20 21 memset(b, 0, sizeof(b)); 22 23 k = 0; b[0] = -1; // 初始化b[0]为-1 24 25 for(i=0; i<N; i++) { 26 27 scanf("%d",&a[i]); 28 29 if(a[i] > b[k]) b[++k] = a[i]; // 大于b[k]的元素直接放在其后面 30 31 else if(a[i] < b[k]) b[Check(a[i])] = a[i]; // 替换大于它的最小的元素 32 33 else ; 34 } 35 36 printf("%d\n",k); // k+1就是对应的最长子序列长度 37 } 38 39 return 0; 40 } 41 42 int Check(int p) // 二分查找 43 { 44 int mid; 45 int temp = k; 46 int l = 0, r = k; 47 48 while(l <= r) { 49 50 mid = (l + r)/2; 51 52 if(b[mid] >= p) { temp = min(mid, temp); r = mid - 1; } 53 54 else l = mid + 1; 55 } 56 57 return temp; 58 }
hoj_10027_优化DP（LIS）,布布扣,bubuko.com
标签：des style c class blog code
原文地址：http://www.cnblogs.com/sayeter/p/3701950.html
