[HDU 4597] Play Game

博弈类DP题，设状态f[i][j][a][b]表示第一堆里的范围在i-j，第二堆的范围在a-b，Alice可以得到的最大值
则有：
dp[i][j][a][b] = max
{
dp[ i + 1 ] [ j ] [ a ] [ b ]
dp[ i ] [ i - 1 ] [ a ] [ b ]
dp[ i ] [ j ] [ a + 1 ] [ b ]
dp[ i ] [ j ] [ a ] [ b - 1 ]
}

code:

#include <bits/stdc++.h>
using namespace std;

int n;
int a[22], b[22];
int f[22][22][22][22];
int sum1[22], sum2[22];

int dfs(int a1, int a2, int b1, int b2)
{
    int &ans = f[a1][a2][b1][b2];

    int now;
    if (a1 > a2)
    {
        now = sum2[b2] - sum2[b1 - 1];
        if (b1 == b2)
            ans = now;
    }
    else if (b1 > b2)
    {
        now = sum1[a2] - sum1[a1 - 1];
        if (a1 == a2)
            ans = now;
    }
    else
    {
        now = sum1[a2] - sum1[a1 - 1] + sum2[b2] - sum2[b1 - 1];
    }

    if (ans != -1)
        return ans;

    ans = 0;
    if (a1 <= a2)
    {
        ans = max(ans, now - min(dfs(a1 + 1, a2, b1, b2), dfs(a1, a2 - 1, b1, b2)));
    }
    if (b1 <= b2)
    {
        ans = max(ans, now - min(dfs(a1, a2, b1 + 1, b2), dfs(a1, a2, b1, b2 - 1)));
    }
    return ans;
}

int main()
{
    // freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &a[i]);
            sum1[i] = sum1[i - 1] + a[i];
        }
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &b[i]);
            sum2[i] = sum2[i - 1] + b[i];
        }

        memset(f, -1, sizeof(f));
        cout << dfs(1, n, 1, n) << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/wyctstf/p/11372420.html