HDU 4925 Apple Tree (Blind)

Find a pattern, plant a tree, or fertilize. Fertilize first, plant trees.

Apple TreeTime limit:2000/1000 MS (java/others) Memory limit:262144/262144 K (java/others)
Total submission (s): 197 Accepted Submission (s): 135


Problem Descriptioni ' ve bought an orchard and decide to plant some apple trees on it. The orchard seems like an N * M two-dimensional map. In each grid, I can either plant a apple tree to get one apple or fertilize the soil to speed up its neighbors ' productio N. When a grid is fertilized, the grid itself doesn ' t produce apples and the number of apples of its four neighbor trees W Ill double (if it exists). For example, a apple tree locates on (x, Y), and (X-1, y), (x, y-1) is fertilized while (x + 1, y), (x, y + 1) is n OT, then I can get four apples from (x, y). Now, I am wondering how many apples I can get on the most in the whole orchard?

Inputthe input contains multiple test cases. The number of test cases T (t<=100) occurs in the first line of input.
For each test case, the integers N, M (1<=n, m<=100) is given in a line, which denote the size of the map.
Outputfor Each test case, you should output the maximum number of apples I can obtain.
Sample Input

22 23 3

Sample Output

832

#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include < iomanip> #include <stdio.h> #include <string> #include <queue> #include <cmath> #include < stack> #include <map> #include <set> #define EPS 1e-12///#define M 1000100#define ll __int64///#define LL Long long///#define INF 0x7ffffff#define inf 0x3f3f3f3f#define PI 3.1415926535898#define Zero (x) ((Fabs (x) <eps)? 0:x)    using namespace Std;const int MAXN = 110;int Mp[maxn][maxn];int main () {int n, m;    int T;    CIN >>T;        while (t--) {scanf ("%d%d", &n, &m);        Memset (MP, 0, sizeof (MP));                    for (int i = 1; I <= N, i++) {if (i%2) {for (int j = 1; j <= m; j + = 2)            MP[I][J] = 1;            } else {for (int j = 2; j <= m; j+= 2) mp[i][j] = 1;      }} LL sum = 0;  for (int i = 1, i <= N; i++) {for (int j = 1; j <= M; j + +) {if (mp[i][j                ]) continue;                int cnt = 0;                if (i-1 >= 1) if (Mp[i-1][j]) cnt + +;                if (i+1 <= N) if (Mp[i+1][j]) cnt + +;                if (j-1 >= 1) if (mp[i][j-1]) cnt++;                if (j+1 <= m) if (mp[i][j+1]) cnt++;            Sum + = (1ll<<cnt);        }} if (sum = = 0) sum++;    cout<<sum<<endl; }}