PKU2385 - Apple Catching (DP)

题意

最多能收到多少苹果。

思路

$dp[i][j][k]$,表示前i秒现在站在j处还剩k次瞬移。

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 = 100 + 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[1010][2][35], arr[1010];
int main()
{
    //ROP;
    int n, l;
    scanf("%d%d", &n, &l);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &arr[i]);
        arr[i]--;
    }
    for (int i = 1; i <= n; i++)
    for (int j = 0; j < 2; j++)
    for (int k = 0; k <= l; k++) dp[i][j][k] = -INF;
    dp[1][arr[1]][l] = 1; dp[1][arr[1]^1][l] = 0;
    for (int i = 2; i <= n; i++)
    {
        for (int j = 0; j < 2; j++)
        {
            for (int k = 0; k <= l; k++)
            {
                dp[i][j][k] = dp[i-1][j][k];
                if (k != l) dp[i][j][k] = max(dp[i][j][k], dp[i-1][j^1][k+1]);
                if (j == arr[i]) dp[i][j][k]++;
            }
        }
    }
    int ans = 0;
    for (int i = 0; i <= l; i++)
    {
        ans = max(ans, dp[n][0][i]);
        ans = max(ans, dp[n][1][i]);
    }
    printf("%d\n", ans);
    return 0;
}

Powered by Jekyll and Theme by solid