乌龟棋
思路
最优值问题,显然可以通过 d p dp dp解决,我们定义 d p [ i ] [ j ] [ k ] [ l ] dp[i][j][k][l] dp[i][j][k][l]表示到达 1 + i ∗ 2 ∗ j + 3 ∗ k + 4 ∗ l 1 + i * 2 * j + 3 * k + 4 * l 1+i∗2∗j+3∗k+4∗l这个点之前已经走过的价值最大的值( i , j , k , l i, j, k, l 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 d p [ i − 1 ] [ j ] [ k ] [ l ] dp[i - 1][j][k][l] dp[i−1][j][k][l]或或者 d p [ i ] [ j − 1 ] [ k ] [ l ] dp[i][j - 1][k][l] dp[i][j−1][k][l]或者 d p [ i ] [ j ] [ k − 1 ] [ l ] dp[i][j][k - 1][l] dp[i][j][k−1][l]或者 d p [ i ] [ j ] [ k ] [ l − 1 ] dp[i][j][k][l - 1] dp[i][j][k][l−1]转移过来,因此我们只需要用四重 f o r for for循环来进行 d p dp dp即可得到我们的最优值,同时输出答案加上点 n n 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;
}