乌龟棋

思路

最优值问题，显然可以通过$dp$解决，我们定义$dp[i][j][k][l]$表示到达$1+i\ast 2\ast j+3\ast k+4\ast l$这个点之前已经走过的价值最大的值($i,j,k,l$分别是走一步，走两步，走三步，走四步的数量)，显然这个点我们可以从$dp[i-1][j][k][l]$或或者$dp[i][j-1][k][l]$或者$dp[i][j][k-1][l]$或者$dp[i][j][k][l-1]$转移过来，因此我们只需要用四重$for$循环来进行$dp$即可得到我们的最优值，同时输出答案加上点$n$的权值即可。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 400;int n, m, num[5], cost[N], dp[45][45][45][45];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read();for(int i = 1; i <= n; i++) {cost[i] = read();}for(int i = 1; i <= m; i++) {num[read()]++;}for(int i = 0; i <= num[1]; i++) {for(int j = 0; j <= num[2]; j++) {for(int k = 0; k <= num[3]; k++) {for(int l = 0; l <= num[4]; l++) {int pos = 1 + i + 2 * j + 3 * k + 4 * l;if(i)   dp[i][j][k][l] = max(dp[i][j][k][l], dp[i - 1][j][k][l] + cost[pos - 1]);if(j)   dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j - 1][k][l] + cost[pos - 2]);if(k)   dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j][k - 1][l] + cost[pos - 3]);if(l)   dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j][k][l - 1] + cost[pos - 4]);}}}}//我们记录的是dp[num[1]][num[2]][num[3]][num[4]]之前的花费，所以答案还要加上这个点的花费。printf("%d\n", dp[num[1]][num[2]][num[3]][num[4]] + cost[n]);return 0;
}