HUD-4925-Apple Tree
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Apple Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1407 Accepted Submission(s): 839
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.
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
22 23 3
Sample Output
832
题意:吃苹果,然后最多走m步。简单说就是从根节点出发走m步所能获得的最大价值。
简单题就不解释了,不过感觉当时用的word例题上的代码不够简洁。。。
代码:
#include<iostream>#include<stdio.h>#include<string.h>#include<vector>#include<math.h>using namespace std;int n,k;int v[105];vector<int>tree[105];int dp[105][205][2];void dfs(int p,int before){for(int i=0;i<tree[p].size();i++) { int next=tree[p][i]; if(next==before) continue; dfs(next,p); for(int j=k;j>=1;j--) { for(int z=1;z<=j;z++) { dp[p][j][0]=max(dp[p][j][0],dp[p][j-z][1]+dp[next][z-1][0]); dp[p][j][0]=max(dp[p][j][0],dp[p][j-z][0]+dp[next][z-2][1]); dp[p][j][1]=max(dp[p][j][1],dp[p][j-z][1]+dp[next][z-2][1]); } } }}int main(){int fa,son;while(cin>>n>>k) { for(int i=0;i<105;i++) tree[i].clear(); for(int i=1;i<=n;i++) { scanf("%d",&v[i]); for(int j=0;j<=k;j++) dp[i][j][0]=dp[i][j][1]=v[i]; } for(int i=0;i<n-1;i++) { scanf("%d%d",&fa,&son); tree[fa].push_back(son); tree[son].push_back(fa); } dfs(1,0); cout<<max(dp[1][k][0],dp[1][k][1])<<endl; }return 0;}
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