Number Triangles

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.

PROGRAM NAME: numtri

INPUT FORMAT

The first line 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.

SAMPLE INPUT (file numtri.in)

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

OUTPUT FORMAT

A single line containing the largest sum using the traversal specified.

SAMPLE OUTPUT (file numtri.out)

30

解体的要点在于，把数据转化成二叉树。然后自底向上计算每个节点的高度。但是注意，由于每个节点同时作为两个节点的子节点（一个是左子节点，一个是右子节点）。

这样可能会造成每个节点计算两次，于是设计每个几点的高度初始值为-1. 当判断高度不会-1时，就不必在计算该节点的高度了。之所以初始值不为0，是因为某个test case全为0.

这样还是会造成重复的递归。

我的解法：

/*
ID: yunleis2
PROG: numtri
LANG: C++
*/

#include<iostream>
#include<fstream>
#include<queue>
using namespace std;

#define max(a,b) (a>b?a:b)

class node
{
public:
 node * left;
 node * right;
 int value;
 int length;
 node(node * l,node * r)
 {
  left=l;
  right=r;
 }
 node ()
 {left=NULL;right=NULL;value=0;length=-1;}

};
int  search( node *root);
void print(node *);
int main()
{
 fstream fin("numtri.in",ios::in);
 int line;
 fin>>line;
 int size=(line+1)*line/2;
 node * root=new node();

 queue<node *> q;
 fin>>root->value;
 q.push(root);
 for(int i=1;i<line;i++)
 {
  node * last=NULL;
  for(int j=1;j<=i;j++)
  {
   node * ptr=q.front();
   q.pop();
   if(j==1)
   {
    ptr->left=new node();
    fin>>ptr->left->value;
    q.push(ptr->left);
   }
   else
   {
    ptr->left=last->right;
   }
   ptr->right=new node();
   fin>>(ptr->right)->value;
   last=ptr;
   
   q.push(ptr->right);
  }
 }
 while(!q.empty())
  q.pop();
 //print(root);
 int length=search(root);
 fstream fout("numtri.out",ios::out);
 fout<<root->length<<endl;
 //system("pause");
 
}
void print(node * root)
{
 if(root!=NULL)
 {
  cout<<root->value;
  print(root->left);
  print(root->right);
 }
}
int  search( node *root)
{
 if(root==NULL)
  return 0;
 if(root->length!=-1)
  return root->length;
 root->length=max(search(root->right),search(root->left))+root->value;
 return root->length;
}

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

We keep track (in the "best" array) of total for the best path ending in a given column of the triangle. Viewing the input, a path through the triangle always goes down or down and to the right. To process a new row, the best path total ending at a given column is the maximum of the best path total ending at that column or the one to its left, plus the number in the new row at that column. We keep only the best totals for the current row (in "best") and the previous row (in "oldbest").

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <assert.h>

#define MAXR 1000

int
max(int a, int b)

{

	return a > b ? a : b;

}

void

main(void){

	int best[MAXR], oldbest[MAXR];

	int i, j, r, n, m;

	FILE *fin, *fout;
	fin = fopen("numtri.in", "r");

	assert(fin != NULL);

	fout = fopen("numtri.out", "w");

	assert(fout != NULL);

	fscanf(fin, "%d", &r);

	for(i=0; i<MAXR; i++)

		best[i] = 0;

	for(i=1; i<=r; i++) {

		memmove(oldbest, best, sizeof oldbest);

		for(j=0; j<i; j++) {

			fscanf(fin, "%d", &n);

			if(j == 0)

				best[j] = oldbest[j] + n;

			else

				best[j] = max(oldbest[j], oldbest[j-1]) + n;

		}

	}

	m = 0;

	for(i=0; i<r; i++)

		if(best[i] > m)

			m = best[i];

	fprintf(fout, "%d\n", m);

	exit(0);
}