2 条题解

  • 0
    @ 2025-4-11 21:17:50

    //gg/g/g/g/g/g/g/g/ #include #include <math.h> #include #include <stdio.h> #include #include <string.h> #include using namespace std; #define LL long long const int N = 2000 + 10; const int INF = 0x3f3f3f3f; int a[N],b[N]; int dp[N][N]; int n; int f() { memcpy(b,a,sizeof(b)); sort(b+1, b+n+1); for(int i = 1 ; i <= n ; i++) { int minn = INF; for(int j = 1 ; j <= n ; j++) { minn = min(minn , dp[i-1][j]); dp[i][j] = minn + abs(a[i] - b[j]); } } int ans = INF; for(int i = 1; i <= n ; i++) { ans = min(ans , dp[n][i]); } return ans; } int main() { cin>> n; for(int i = 1 ; i <= n ; i++) cin >> a[i]; int ans = f(); reverse(a+1,a+n+1); ans = min(ans, f()); cout << ans << endl; }

    信息

    ID
    184
    时间
    1000ms
    内存
    128MiB
    难度
    9
    标签
    递交数
    7
    已通过
    7
    上传者