华中农业大学第五届程序设计大赛网络同步赛-G
G. Sequence Number
In Linear algebra, we have learned the definition of inversion number: Assuming A is a ordered set with n numbers ( n > 1 ) which are different from each other. If exist positive integers i , j, ( 1 ≤ i < j ≤ n and A[i] > A[j]), <a[i], a[j]=""> is regarded as one of A’s inversions. The number of inversions is regarded as inversion number. Such as, inversions of array <2,3,8,6,1> are <2,1>, <3,1>, <8,1>, <8,6>, <6,1>,and the inversion number is 5.
Similarly, we define a new notion —— sequence number, If exist positive integers i, j, ( 1 ≤ i ≤ j ≤ n and A[i] <= A[j], <a[i], a[j]=""> is regarded as one of A’s sequence pair. The number of sequence pairs is regarded as sequence number. Define j – i as the length of the sequence pair.
Now, we wonder that the largest length S of all sequence pairs for a given array A.
Input
There are multiply test cases. In each case, the first line is a number N(1<=N<=50000 ), indicates the size of the array, the 2th ~n+1th line are one number per line, indicates the element Ai (1<=Ai<=10^9) of the array.
Output
Output the answer S in one line for each case.
Sample Input
5 2 3 8 6 1
Sample Output
3
暴力+剪枝
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 6 using namespace std; 7 8 const int N = 50005; 9 int A[N], dp[N]; 10 11 int main() 12 { 13 int n; 14 while(scanf("%d", &n) != EOF) 15 { 16 for(int i = 0; i < n; i++) 17 { 18 scanf("%d", &A[i]); 19 } 20 int len = 0; 21 for(int i = 0; i < n; i++) 22 { 23 for(int j = n-1; j > i; j--){ 24 if(j-i < len)break; 25 if(A[i] <= A[j]){ 26 if(j-i > len)len = j-i; 27 } 28 } 29 if(n-1-i < len)break; 30 } 31 printf("%d\n", len); 32 } 33 return 0; 34 }