P1541 [NOIP2010 提高组] 乌龟棋

合集 - 动态规划(7)

1.P1020 [NOIP1999 提高组] 导弹拦截2024-10-172.P1439 【模板】最长公共子序列2024-10-193.P1435 [IOI2000] 回文字串2024-10-19

4.P1541 [NOIP2010 提高组] 乌龟棋2024-10-19

5.P1004 [NOIP2000 提高组] 方格取数2024-10-196.P4933 大师2024-10-197.P1091 [NOIP2004 提高组] 合唱队形2024-10-19

收起

dp[a][b][c][d]表示走了a+b2+c3+d*4步的当前的最大值，状态转移方程就出来了。

点击查看代码

#include <iostream>
#include <stack>
#include <cmath>
#include <algorithm>
#include <set>
#include <vector>
#include <climits>
#include <string.h>
#include <map>
#include <queue>
#include <list>
#include <cmath>
#include <iomanip> 
#define int long long 
#define ios ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define lc u<<1
#define rc u<<1|1
#define gcd __gcd
#define double long double
#define endl "\n"
#define INF LLONG_MAX
#define mod  1000000007
#define N 40
const double PI = 3.14159265358979323846;
using namespace std;
int n, m, num[355], g[N],dp[N][N][N][N];
signed main()
{
    ios;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)cin >> num[i];
    for (int j = 1; j <= m; j++)
    {
        int x; cin >> x;
        g[x]++;
    }
    dp[0][0][0][0] = num[1];
    for (int a = 0; a <= g[1]; a++)
    {
        for (int b = 0; b <= g[2]; b++)
        {
            for (int c = 0; c <= g[3]; c++)
            {
                for (int d = 0; d <= g[4]; d++)
                {
                    int r = 1 + a + b * 2 + c * 3 + d * 4;
                    if (a != 0)dp[a][b][c][d] = max(dp[a][b][c][d], dp[a - 1][b][c][d] + num[r]);
                    if (b != 0)dp[a][b][c][d] = max(dp[a][b][c][d], dp[a][b - 1][c][d] + num[r]);
                    if (c != 0)dp[a][b][c][d] = max(dp[a][b][c][d], dp[a][b][c - 1][d] + num[r]);
                    if (d != 0)dp[a][b][c][d] = max(dp[a][b][c][d], dp[a][b][c][d - 1] + num[r]);
                }

            }

        }
    }
    cout << dp[g[1]][g[2]][g[3]][g[4]] << endl;
    return 0;
}