题意：n*m网格中种苹果，每个网格要么施肥，要么种一个苹果，每个种苹果的格子，如果它的上下左右有各自有施肥的话，每有一个，苹果数量*2，求怎么种使得苹果数量最多。
思路：交叉种植，即黑白染色法可得到最优解。注意特判当n=m=1时的情况。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 105;
int n, m;
int g[MAXN][MAXN];
int deal(int x, int y) {
int sum = 1;
if (g[x][y - 1] == 1)
sum *= 2;
if (g[x][y + 1] == 1)
sum *= 2;
if (g[x - 1][y] == 1)
sum *= 2;
if (g[x + 1][y] == 1)
sum *= 2;
return sum;
}
int main() {
int cas;
scanf("%d", &cas);
while (cas--) {
scanf("%d%d", &n, &m);
if (n == 1 && m == 1) {
printf("1\n");
continue;
}
memset(g, 0, sizeof(g));
int flag = 1;
if (m % 2 == 1) {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
g[i][j] = flag;
flag = -flag;
}
}
else {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
g[i][j] = flag;
flag = -flag;
}
flag = -flag;
}
}
long long ans = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
if (g[i][j] == -1)
ans += deal(i, j);
printf("%lld\n", ans);
}
return 0;
}