最长递增子序列

Posted 2020-12-27 banyouxia

1037: I want you!

Description

Given an array Ai(0<=i<n), its length is n; We want to find an another array Bi , which is 0 or 1,and its length is n too;

Besides, Ai and Bi meets the following conditions:

If neither A[i]*B[i] nor A[j]*B[j] equals to 0, then A[i]*B[i] < A[j]*B[j];(0<=i<j<n)

Now , we want to know the maximum of ∑Bi(0<=i<n) you can get.

Input

The input consists of multiple test cases。For each test case ,the first line contains one integer n (1<n<=300).Next line contains n integers.

Output

For each case, output the maximum of ∑Bi.

Sample Input

6

1 2 3 4 5 6

4

3 2 3 6

Sample Output

6

3

思路：求最长递增子序列，注意求LIS时要把数组中0去掉单独处理，输出LIS长度+0的个数即为答案。

代码如下：

#include<bits/stdc++.h>
using namespace std;
#define maxn 300 + 10
#define INF 10000000000
#define FOR(i,x,y) for(int i = x; i < y; i++)
#define FORD(j,y,x) for(int j = y; j >= x; j--)
#define mset(x,v) memset(x,v,sizeof(x))
int a[maxn];
int b[maxn];
//数组a的最长递增子序列，时间复杂度O(n^2)
int LIS(int *a,int len){
    int maxlen[len];
    FOR(i,0,len) maxlen[i] = 1;
    FOR(j,1,len){
        FOR(i,0,j){
            if(a[j] > a[i] && maxlen[j] < maxlen[i] + 1){
                maxlen[j] = maxlen[i] + 1;
            }
        }
    }
    int max = 0;
    FOR(i,0,len){
        if(maxlen[i] > max) max = maxlen[i];
    }
    return max;
}
int main(){
    int n;
    while(~scanf("%d",&n)){
        int c = 0;
        int j = 0;
        for(int i = 0;i < n;i++){
            scanf("%d",&a[i]);
            if(a[i]==0) {
                c++;
            }else{
                b[j++] = a[i];//除去0的数组b
            }
        }
        int ans = LIS(b,j);//最长递增子序列的长度
        printf("%d ",ans + c);
    }
    return 0;
}