HRBUST 2010简单dp+最长递减子序列

Posted 2020-11-02 tennant

题目：所谓二等队形就是从大到小依次排列，即对于数列a，二等队形为任意a【i】满足：a【i】>a【i+1】。现在给出一个长度为n的数列，从中最少去除多少个数可使数列变成二等队形数列。

做法和递增的没啥差别

#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#define MAX 1000+10
#define INF 0x3f3f3f3f
int n;
int dp[MAX];

using namespace std;

int main(void) {
    int a[MAX];
    while (~scanf("%d", &n)) {
        for (int i = 1; i <= n; i++)
            dp[i] = 1;
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);

        a[0] = INF;
        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (a[i] < a[j]) {
                    dp[i] = max(dp[j] + 1, dp[i]);

                }

            }

        }
        int maxx = -INF;
        for (int i = 1; i <= n; i++) {
            maxx = max(dp[i], maxx);

        }
        printf("%d\n", n - maxx);

    }
    return 0;
}