2017 ICPC 南宁区域赛 I Rake It In

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题目链接：

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题意描述：

Alice和Bob在玩一种名为“Rake It In”的游戏，起初有一个4*4的棋盘，每一格为一个1~10的整数，两人轮流行动，各自k次，行动者选择棋盘中某一个2*2的区域，将这四个元素求和，加到最终答案中，并将四个元素按逆时针旋转90度，Alice先行。Alice的目标是最大化最后的答案，Bob相反。请写一个程序计算，在两人都最聪明的情况下，最终答案为多少。

解题思路：

对于每一次旋转或者求和，在4*4的棋盘下选出2*2的区域总共只有9种。

由于k<=3，t<=200，可以想到，在最坏的情况下，暴力计算出每一层的最优解，一共也只有6层，时间最多为200*（9^6）。

因此采用贪心+暴力DFS的方法，暴力递归出每一层的最优解，进而求取最终答案，代码如下：

复制代码

#include<cstdio>
#include<iostream>
using namespace std;
#define INF 0x3f3f3f3f

int a[5][5];
int k;

int cal(int t)
{
    int i = (t+2)/3;
    int j = (t-1)%3+1;
    return a[i][j] + a[i+1][j] + a[i][j+1] + a[i+1][j+1];
}

void rotat(int t)
{
    int i = (t+2)/3;
    int j = (t-1)%3+1;
    int a1 = a[i][j];
    int a2 = a[i+1][j];
    int a3 = a[i][j+1];
    int a4 = a[i+1][j+1];
    a[i][j] = a3;
    a[i+1][j] = a1;
    a[i+1][j+1] = a2;
    a[i][j+1] = a4;
}

void rerotat(int t)
{
    int i = (t+2)/3;
    int j = (t-1)%3+1;
    int a1 = a[i][j];
    int a2 = a[i+1][j];
    int a3 = a[i][j+1];
    int a4 = a[i+1][j+1];
    a[i][j] = a2;
    a[i+1][j] = a4;
    a[i+1][j+1] = a3;
    a[i][j+1] = a1;
}

int dfs(int t)
{
    if(t == 2*k)
    {
        int ans = INF;
        for(int i=1; i<=9; i++)
            ans = min(ans, cal(i));
        return ans;
    }
    else if(t%2 == 0)
    {
        int ans = INF;
        for(int i=1; i<=9; i++)
        {
            rotat(i);
            int temp = dfs(t+1);
            ans = min(ans, temp+cal(i));
            rerotat(i);
        }
        return ans;
    }
    else
    {
        int ans = 0;
        for(int i=1; i<=9; i++)
        {
            rotat(i);
            int temp = dfs(t+1);
            ans = max(ans, temp+cal(i));
            rerotat(i);
        }
        return ans;
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &k);
        for(int i=1; i<=4; i++)
            for(int j=1; j<=4; j++)
                scanf("%d", &a[i][j]);
        printf("%dn", dfs(1));
    }
    return 0;
}

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#include<cstdio>
#include<iostream>
using namespace std;
#define INF 0x3f3f3f3f

int a[5][5];
int k;

int cal(int t)
{
    int i = (t+2)/3;
    int j = (t-1)%3+1;
    return a[i][j] + a[i+1][j] + a[i][j+1] + a[i+1][j+1];
}

void rotat(int t)
{
    int i = (t+2)/3;
    int j = (t-1)%3+1;
    int a1 = a[i][j];
    int a2 = a[i+1][j];
    int a3 = a[i][j+1];
    int a4 = a[i+1][j+1];
    a[i][j] = a3;
    a[i+1][j] = a1;
    a[i+1][j+1] = a2;
    a[i][j+1] = a4;
}

void rerotat(int t)
{
    int i = (t+2)/3;
    int j = (t-1)%3+1;
    int a1 = a[i][j];
    int a2 = a[i+1][j];
    int a3 = a[i][j+1];
    int a4 = a[i+1][j+1];
    a[i][j] = a2;
    a[i+1][j] = a4;
    a[i+1][j+1] = a3;
    a[i][j+1] = a1;
}

int dfs(int t)
{
    if(t == 2*k)
    {
        int ans = INF;
        for(int i=1; i<=9; i++)
            ans = min(ans, cal(i));
        return ans;
    }
    else if(t%2 == 0)
    {
        int ans = INF;
        for(int i=1; i<=9; i++)
        {
            rotat(i);
            int temp = dfs(t+1);
            ans = min(ans, temp+cal(i));
            rerotat(i);
        }
        return ans;
    }
    else
    {
        int ans = 0;
        for(int i=1; i<=9; i++)
        {
            rotat(i);
            int temp = dfs(t+1);
            ans = max(ans, temp+cal(i));
            rerotat(i);
        }
        return ans;
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &k);
        for(int i=1; i<=4; i++)
            for(int j=1; j<=4; j++)
                scanf("%d", &a[i][j]);
        printf("%dn", dfs(1));
    }
    return 0;
}