题目链接:https://nanti.jisuanke.com/t/A1538
题解:因为k只有3,所以最多取6次,每次有9种选取可能,所以直接dfs选取即可
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int b[6][6];
int k;
int dfs(int u) {
if(u > 2 * k) return 0;
int ans = 1e9, cnt;
int tmp;
if(u & 1) {
ans = 0;
for(int i = 1; i <= 3; i++) {
for(int j = 1; j <= 3; j++) {
cnt = b[i][j] + b[i + 1][j] + b[i][j + 1] + b[i + 1][j + 1];
tmp = b[i][j];
b[i][j] = b[i][j + 1];
b[i][j + 1] = b[i + 1][j + 1];
b[i + 1][j + 1] = b[i + 1][j];
b[i + 1][j] = tmp;
ans = max(ans, cnt + dfs(u + 1));
tmp = b[i][j];
b[i][j] = b[i + 1][j];
b[i + 1][j] = b[i + 1][j + 1];
b[i + 1][j + 1] = b[i][j + 1];
b[i][j + 1] = tmp;
}
}
} else {
for(int i = 1; i <= 3; i++) {
for(int j = 1; j <= 3; j++) {
cnt = b[i][j] + b[i + 1][j] + b[i][j + 1] + b[i + 1][j + 1];
tmp = b[i][j];
b[i][j] = b[i][j + 1];
b[i][j + 1] = b[i + 1][j + 1];
b[i + 1][j + 1] = b[i + 1][j];
b[i + 1][j] = tmp;
ans = min(ans, cnt + dfs(u + 1));
tmp = b[i][j];
b[i][j] = b[i + 1][j];
b[i + 1][j] = b[i + 1][j + 1];
b[i + 1][j + 1] = b[i][j + 1];
b[i][j + 1] = tmp;
}
}
}
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", &b[i][j]);
printf("%d\n", dfs(1));
}
return 0;
}