51Nod－1270－数组的最大代价

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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;}

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