Number Triangles

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

这题不难，就是每个元素取与不取，这一层的某个元素最大路径由上一层的左右决定  ps：sort函数对二维数组无效这个没考虑到。

/*
ID: des_jas1
PROG: numtri
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <string.h>
#include <algorithm>
//#define fin cin
//#define fout cout
#define max(a,b) ((a)>(b)?(a):(b))
using namespace std;

const int MAX=1000+5;
int R,c[MAX][MAX];

void swap(int &a,int &b)
{
	int tp;
	tp=a;
	a=b;
	b=tp;
}

int main() 
{
	ofstream fout ("numtri.out");
    ifstream fin ("numtri.in");
	int i,j,a;
	fin>>R;
	memset(c,0,sizeof(c));
	for(i=1;i<=R;i++)
		for(j=1;j<=i;j++)
		{
			fin>>a;
			c[i][j]=a+max(c[i-1][j-1],c[i-1][j]);
		}
	a=0;
	for(j=1;j<=R;j++)
	{
		if(c[R][j]>a)
			a=c[R][j];
	}
	fout<<a<<endl;
	fout.close();
	fin.close();
    return 0;
}