Pku1163 the Triangle （动态规划） 数字三角形

H - The Triangle

Time Limit:1000MS    Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

 

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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

对于一个有数字组成的二叉树，求由叶子到根的一条路径，使数字和最大，如：

                                  7

                                3    8

                              8   1   0

                            2   7   4   4

                          4   5   2   6   5

这个是经典的动态规划，也是最最基础、最最简单的动态规划，典型的多段图。思路就是建立一个数组，由上向下动态规划，保存上节点到叶节点的最大值

记忆化搜索法：

#include<stdio.h> //0ms AC#include<string.h>int map[105][105],d[105][105],n;int dis(int i,int j){if(d[i][j]>=0) return d[i][j];return d[i][j]=map[i][j]+(i==n ? 0 : (dis(i+1,j)>dis(i+1,j+1)?dis(i+1,j):dis(i+1,j+1)));}int main(){int i,j;scanf("%d",&n);memset(d,-1,sizeof(d));for(i=1;i<=n;i++)for(j=1;j<=i;j++)scanf("%d",&map[i][j]);printf("%d\n",dis(1,1));return 0;}

好吧，还有个递推计算法，从下往上的

#include<stdio.h> //16ms ACint map[105][105],d[105][105];int main(){int n,i,j;scanf("%d",&n);for(i=0;i<n;i++)for(j=0;j<=i;j++)scanf("%d",&map[i][j]);for(i=0;i<n;i++) d[n-1][i]=map[n-1][i];for(i=n-2;i>=0;i--)for(j=0;j<=i;j++)d[i][j]=map[i][j]+(d[i+1][j]>d[i+1][j+1]?d[i+1][j]:d[i+1][j+1]);printf("%d\n",d[0][0]);return 0;}