ACM模版

描述

题解

动态规划，极端考虑法，每个A[i]要么取1，要么取B[i]。
状态转移方程也很好推（dp[i][j]:j->0表示A[i]取1，j->1表示A[i]取B[i]）:

dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + B[i - 1] - 1);
dp[i][1] = max(dp[i - 1][0] + B[i] - 1, dp[i - 1][1] + abs(B[i] - B[i - 1]));

代码

One：

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int MAXN = 5e4 + 10;

int B[MAXN];
int dp[MAXN][2] = {
  0};  // dp[i][j]:j->0表示A[i]取1，j->1表示A[i]取B[i]

int main(int argc, const char * argv[])
{
    int N;
    cin >> N;

    for (int i = 1; i <= N; i++)
    {
        scanf("%d", B + i);
    }

    for (int i = 2; i <= N; i++)
    {
        dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + B[i - 1] - 1);
        dp[i][1] = max(dp[i - 1][0] + B[i] - 1, dp[i - 1][1] + abs(B[i] - B[i - 1]));
    }

    std::cout << max(dp[N][0], dp[N][1]) << "\n";
    return 0;
}

Two：

// 空间优化，滚动数组
#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int MAXN = 5e4 + 10;

int B[MAXN];

int main()
{
    int N;
    cin >> N;

    for (int i = 0; i < N; i++)
    {
        scanf("%d", B + i);
    }

    int dpx[2], dp[2] = {
  0, 0};
    for (int i = 1; i < N; i++)
    {
        dpx[0] = max(dp[0], dp[1] + B[i - 1] - 1);
        dpx[1] = max(dp[0] + B[i] - 1, dp[1] + abs(B[i] - B[i - 1]));
        dp[0] = dpx[0];
        dp[1] = dpx[1];
    }

    cout << max(dp[0], dp[1]);

    return 0;
}