【动态规划】The Triangle

问题 E: 【动态规划】The Triangle

时间限制: 1 Sec  内存限制: 128 MB
提交: 24  解决: 24
[提交][状态][讨论版]

题目描述

7
3 8
8 1 0
2 7 4 4
4 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.

输入

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.

输出

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

样例输入

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

样例输出

30解题思路：首先从上往下是i到n-1，从左往右是j到i；　　要从上往下加，先看最上面的7，可以加3等于10，可以加8等于15；　　然后看第三行当j等于0，只能加到它右上方那个上，当j=i，只能加到左上方那个数上。　　当j!=0或i的时候，可以加到左上方，可以加到右上方，但要求最后的和最大，所以要加到和大的那一个上面。　　并且加到大的那一个上这一决策对之后的没有影响，无后效性。　　状态转移方程为：sum[i][j]=max(sum[i-1][j],sum[i-1][j-1])+a[i][j];　　从上到下，从左到右依次遍历，最后一行其中一个数上会有最大值。　　便利最后一行sum[i]，找出最大值。

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int a[101][101];
    int sum[101][101];
    int n;
    int maxx;
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            for(int j=0;j<i+1;j++){
                scanf("%d",&a[i][j]);
            }
        }
        sum[0][0]=a[0][0];
        for(int i=1;i<n;i++){
            for(int j=0;j<i+1;j++){
                if(j==0){
                    sum[i][j]=sum[i-1][j]+a[i][j];
                }
                if(j==i){
                    sum[i][j]=sum[i-1][i-1]+a[i][j];
                }
                sum[i][j]=max(sum[i-1][j],sum[i-1][j-1])+a[i][j];
            }
        }
        maxx=0;
        for(int i=0;i<n;i++){
            if(maxx<sum[n-1][i]){
                maxx=sum[n-1][i];
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}