PKU 2181 - Jumping Cows (DP)
题意
跳奇数步加上能量,偶数步减去,问最大能量。
思路
背包。如果某个阶梯不跳的话就直接继承上一个相同特性的阶梯值。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 150000 + 10;
const int MOD = 9901;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
int dp[MAXN][2], arr[MAXN];
int main()
{
//ROP;
//odd increase
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
for (int i = 1; i <= n; i++)
{
dp[i][0] = max(dp[i-1][0], dp[i-1][1] - arr[i]);
dp[i][1] = max(dp[i-1][1], dp[i-1][0] + arr[i]);
}
printf("%d\n", max(dp[n][0], dp[n][1]));
return 0;
}