NC16590 乌龟棋(记忆化搜索)

题目链接

题意：






题解：

AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;
const double eps = 1e-6;
const double PI = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 120 + 5;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

int v[maxn], cnt[10], dp[N][N][N][N], n, m;
int dfs(int p, int a, int b, int c, int d) {
    if (p == n) return v[p];
    if (dp[a][b][c][d]) return dp[a][b][c][d];
    int ans = 0;
    if (a < cnt[1]) ans = max(dfs(p + 1, a + 1, b, c, d), ans);
    if (b < cnt[2]) ans = max(dfs(p + 2, a, b + 1, c, d), ans);
    if (c < cnt[3]) ans = max(dfs(p + 3, a, b, c + 1, d), ans);
    if (d < cnt[4]) ans = max(dfs(p + 4, a, b, c, d + 1), ans);
    return dp[a][b][c][d] = ans + v[p];
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> v[i];
    for (int i = 1, x; i <= m; i++) {
        cin >> x;
        cnt[x]++;
    }
    cout << dfs(1, 0, 0, 0, 0);
    return 0;
}