FZU1004-Number Triangle经典动归题，核心思路及代码优化

Posted 2020-06-25

 

Problem 1004 Number Triangle

Accept: 2230    Submit: 5895
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

7        3   8      8  1  0    2  7  4  4  4  5  2   6  5

In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.

 Input

There are multiple test cases.The first line of each test case contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.

 Output

Print a single line containing the largest sum using the traversal specified for each test case.

 Sample Input

5

7

3 8

8 1 0

2 7 4 4

4 5 2 6 5

 Sample Output

30

   很明显的一道动归题，类似于求最大路径，方法有很多，搜索啊，递归啊，，，但用动归是最简单的，动归思路有两种：

   1：

用二维数组存放数字三角形；

D( r, j) : 第r行第 j 个数字(r,j从1 开始算)

MaxSum(r, j) : 从D(r,j)到底边的各条路径中最佳路径的数字之和，问题是就求MaxSum(1,1)；
典型的递归问题
D(r, j)出发，下一步只能走D(r+1,j)或者D(r+1, j+1)。故对于N行的三角形：
if ( r == N)
MaxSum(r,j) = D(r,j)
else
MaxSum( r, j) = Max{ MaxSum(r＋1,j), MaxSum(r+1,j+1) } + D(r,j)；

2： 如果每算出一个MaxSum(r,j)就保存起来，下次用到其值的时候直接取用，则可免去重复计算。

那么可以用 O(n2)时间完成计算。因为三角形的数字总数是 n(n+1)/2；这时，递归转化为递推；

 

 下面呈现两种思路代码：

  1：超时代码：       

#include <iostream>
#include <algorithm>
#define MAX 101
using namespace std;
int D[MAX][MAX];
int n;
int MaxSum(int i, int j){
 if(i==n)
return D[i][j];
 int x = MaxSum(i+1,j);
 int y = MaxSum(i+1,j+1);
 return max(x,y)+D[i][j];
}
int main(){
 int i,j;
 cin >> n;
 for(i=1;i<=n;i++)
for(j=1;j<=i;j++)
 cin >> D[i][j];
 cout << MaxSum(1,1) << endl;
}
如果采用递规的方法，深度遍历每条路径，存在大量重复计算。则时间复杂度为 2n, 对于 n = 100行，肯定超时。

 

  2：记忆递归

#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 101
int D[MAX][MAX]; int n;
int maxSum[MAX][MAX];
int MaxSum(int i, int j){
 if( maxSum[i][j] != -1 )
  return maxSum[i][j];
 if(i==n) maxSum[i][j] = D[i][j];
 else {
 int x = MaxSum(i+1,j);
  int y = MaxSum(i+1,j+1);
  maxSum[i][j] = max(x,y)+ D[i][j];
 }
 return maxSum[i][j];
}
int main(){
 int i,j;
 cin >> n;
 for(i=1;i<=n;i++)
for(j=1;j<=i;j++) {
 cin >> D[i][j];
 maxSum[i][j] = -1;
}
 cout << MaxSum(1,1) << endl;
}

 

  3：

#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 101
int D[MAX][MAX]; int n;
int maxSum[MAX][MAX];
int main() {
 int i,j;
 cin >> n;
 for(i=1;i<=n;i++)
for(j=1;j<=i;j++)
 cin >> D[i][j];
 for( int i = 1;i <= n; ++ i )
 maxSum[n][i] = D[n][i];
 for( int i = n-1; i>= 1; --i )
for( int j = 1; j <= i; ++j )
 maxSum[i][j] = max(maxSum[i+1][j],maxSum[i+1][j+1]) + D[i][j]
 cout << maxSum[1][1] << endl;
}
人人为我”递推型动归程序

 

 4：当然了，3中的代码也可以空间优化，从下往上找

    

#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 101
int D[MAX][MAX];
int n; int * maxSum;
int main(){
 int i,j;
 cin >> n;
 for(i=1;i<=n;i++)
for(j=1;j<=i;j++)
 cin >> D[i][j];
 maxSum = D[n]; //maxSum指向第n行
 for( int i = n-1; i>= 1; --i )
for( int j = 1; j <= i; ++j )
 maxSum[j] = max(maxSum[j],maxSum[j+1]) + D[i][j];
 cout << maxSum[1] << endl;
}
如果看不懂简单手推模拟一遍你就明白了，不难。。和3中代码类似，只是将二维转化为一维了；