題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4925

Apple Tree

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 188????Accepted Submission(s): 129


Problem Description I’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 an apple tree to get one apple or fertilize the soil to speed up its neighbors’ production. When a grid is fertilized, the grid itself doesn’t produce apples but the number of apples of its four neighbor trees will double (if it exists). For example, an apple tree locates on (x, y), and (x - 1, y), (x, y - 1) are fertilized while (x + 1, y), (x, y + 1) are not, then I can get four apples from (x, y). Now, I am wondering how many apples I can get at most in the whole orchard?

Input The input contains multiple test cases. The number of test cases T (T<=100) occurs in the first line of input.
For each test case, two integers N, M (1<=N, M<=100) are given in a line, which denote the size of the map.
Output For each test case, you should output the maximum number of apples I can obtain.
Sample Input 2 2 2 3 3
Sample Output 8 32
Source 2014 Multi-University Training Contest 6
Recommend hujie???|???We have carefully selected several similar problems for you:??4929?4928?4926?4924?4923?
Statistic?|?Submit?|?Discuss?|?Note

簽道題，沒啥好說的，黑白染色的方法是最優的，特判1*1的情況

#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<string> #include<vector> #include<algorithm> #include<queue> #include<stack> #include<set> #include<map> using namespace std; #define CLR(A) memset(A,0,sizeof(A)) int A[110][110]; int main(){int T,m,n;cin>>T;while(T--){cin>>n>>m;if(n==1 && m==1){cout<<1<<endl;continue;}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)A[i][j]=1;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(A[i][j]==1){A[i-1][j]<<=1;A[i+1][j]<<=1;A[i][j-1]<<=1;A[i][j+1]<<=1;}}long long sum=0;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(A[i][j]!=1){sum+=A[i][j];}}cout<<sum<<endl;}return 0; }

轉載于:https://www.cnblogs.com/wzzkaifa/p/6901870.html

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