poj1163——The Triangle

原题：

Description

73   88   1   02   7   4   44   5   2   6   5(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers 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. 

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

573 88 1 0 2 7 4 44 5 2 6 5

Sample Output

30

 

分析：

动态规划初涉~

源码：

#include<string.h>#include<stdio.h>#define N 200int d[N][N];int n;int a[N][N];int p(int r,int j){    if(r==n)        return d[r][j];    if(a[r+1][j]==0)//如果p(r+1,j)没有计算过        a[r+1][j]=p(r+1,j);    if(a[r+1][j+1]==0)//如果p(r+1,j+1)没有计算过        a[r+1][j+1]=p(r+1,j+1);    if(a[r+1][j]>a[r+1][j+1])        return a[r+1][j]+d[r][j];    else return a[r+1][j+1]+d[r][j];//可以省去else？}int main(){    int m;    scanf("%d",&n);    //将a全部置成0；表示开始所有的a（r，j）都没有算过    memset(a,0,sizeof(a));    for(int i=1; i<=n; i++)    {        for(int j=1; j<=i; j++)        {            scanf("%d",&d[i][j]);        }    }    printf("%d",p(1,1));    return 0;}

 附：未优化（递归）代码——

#include<string.h>#include<stdio.h>#define N 200int d[N][N];int n;int p(int r,int k){    if(r==n)        return d[r][k];    if(p(r+1,k)>p(r+1,k+1))        return p(r+1,k)+d[r][k];    else return p(r+1,k+1)+d[r][k];}int main(){    int m;    scanf("%d",&n);    for(int i=1; i<=n; i++)    {        for(int j=1; j<=i; j++)        {            scanf("%d",&d[i][j]);        }    }    printf("%d",p(1,1));    return 0;}