1.5.1---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

典型的动态规划题目。

/*
ID:******
PROG:numtri
LANG:C++
*/
#include <iostream>
#include <fstream>
#include <cstring>
using namespace std;
#define max 1001
int st[max][max];//存储数塔
int dp(int r){//动态规划


for(int k = 1;k <= r;k ++)
if(st[r+1][k] >= st[r+1][k+1])
st[r][k] = st[r][k] + st[r+1][k];
else
st[r][k] = st[r][k] + st[r+1][k+1];
if(r == 1)
return st[r][r];//退出条件
else
  dp(r - 1);//由下到上
}
int main(){
int R;
ofstream fout ("numtri.out");
ifstream fin ("numtri.in");
fin>>R;
for(int i = 1;i <= R; i++)
for(int j=1; j <= i;j ++)
fin>>st[i][j];
if(R>1)//这里要注意，TEST3的数据为1 0 真坑
fout<<dp(R-1)<<endl;
else
fout<<st[R][R]<<endl;
return 0;


}