hdu-2084-数塔-dp-java
数塔
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 53741 Accepted Submission(s): 31567
Problem Description
在讲述DP算法的时候,一个经典的例子就是数塔问题,它是这样描述的:
有如下所示的数塔,要求从顶层走到底层,若每一步只能走到相邻的结点,则经过的结点的数字之和最大是多少?
![hdu-2084-数塔-dp-java hdu-2084-数塔-dp-java](/default/index/img?u=aHR0cHM6Ly9waWFuc2hlbi5jb20vaW1hZ2VzLzY3My83OWI2NTA5MjczNGQ2YTI2M2E5MjgwZjViMzMyYzViOS5KUEVH)
已经告诉你了,这是个DP的题目,你能AC吗?
有如下所示的数塔,要求从顶层走到底层,若每一步只能走到相邻的结点,则经过的结点的数字之和最大是多少?
已经告诉你了,这是个DP的题目,你能AC吗?
Input
输入数据首先包括一个整数C,表示测试实例的个数,每个测试实例的第一行是一个整数N(1 <= N <= 100),表示数塔的高度,接下来用N行数字表示数塔,其中第i行有个i个整数,且所有的整数均在区间[0,99]内。
Output
对于每个测试实例,输出可能得到的最大和,每个实例的输出占一行。
Sample Input
1
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output
30
Source
2006/1/15 ACM程序设计期末考试
解题思路:
额好奇的我写了个递归搜索果然超时(手动笑哭)然后老老实实的用dp写 emmm 从底自上dp 没啥可说的
import java.util.Scanner;
public class Main {
private static int n;
private static int[][] arr;
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
n = scanner.nextInt();
arr = new int [n+1][n+1];
int [][] dp = new int [n+1][n+1];
for (int j = 1; j < arr.length-1; j++) {
for (int j2 = 1; j2 <= j; j2++) {
arr[j][j2] = scanner.nextInt();
}
}
for (int j = 1; j < dp.length; j++) {
arr[n][j] = scanner.nextInt();
dp[n][j] = arr[n][j];
}
for (int j = n-1; j >= 1; j--) {
for (int j2 = 1; j2 <= j; j2++) {
dp[j][j2] = Math.max(dp[j+1][j2], dp[j+1][j2+1]) + arr[j][j2];
}
}
System.out.println(dp[1][1]);
}
}
}
解题思路:
额好奇的我写了个递归搜索果然超时(手动笑哭)然后老老实实的用dp写 emmm 从底自上dp 没啥可说的
import java.util.Scanner;
public class Main {
private static int n;
private static int[][] arr;
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
n = scanner.nextInt();
arr = new int [n+1][n+1];
int [][] dp = new int [n+1][n+1];
for (int j = 1; j < arr.length-1; j++) {
for (int j2 = 1; j2 <= j; j2++) {
arr[j][j2] = scanner.nextInt();
}
}
for (int j = 1; j < dp.length; j++) {
arr[n][j] = scanner.nextInt();
dp[n][j] = arr[n][j];
}
for (int j = n-1; j >= 1; j--) {
for (int j2 = 1; j2 <= j; j2++) {
dp[j][j2] = Math.max(dp[j+1][j2], dp[j+1][j2+1]) + arr[j][j2];
}
}
System.out.println(dp[1][1]);
}
}
}