[DP] POJ - 1163 The Triangle

最新推荐文章于 2022-03-16 18:28:20 发布

原创最新推荐文章于 2022-03-16 18:28:20 发布 · 236 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj #dp

算法_动态规划专栏收录该内容

4 篇文章

订阅专栏

本文探讨了如何解决三角形路径最大和问题，即在给定的三角形中找到从顶点到底部路径上的数字之和的最大值。通过动态规划的方法，文章提供了两种不同的实现方案，并附带了简洁高效的代码示例。

The Triangle
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 51206 Accepted: 31010

Description

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.
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

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

30

题意:输入一个n层的三角形，求从第1层到第n层的所有路线中，权值之和最大的路线。

规定：第i层的某个数只能连线走到第i+1层中与它位置相邻的两个数中的一个。

//已AC代码
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
using namespace std;

int main()
{
    int t;
    int ans=0;
    int a[188][188];
    int dp[188][188];
    memset(a,0,sizeof(a));
    memset(dp,0,sizeof(dp));
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        for(int j=1;j<=i;j++)
        {
            cin>>a[i][j];
        }
    }
    for(int i=1;i<=t;i++)
    {
        for(int j=1;j<=i;j++)
        {
            if(i==1&&j==1)
            {
                dp[i][j]=a[1][1];
            }
            else if(j==1)
            {
                dp[i][j]=dp[i-1][j]+a[i][j];
            }
            else if(j==i)
            {
                dp[i][j]=dp[i-1][j-1]+a[i][j];
            }
            else
                dp[i][j]=a[i][j]+max(dp[i-1][j],dp[i-1][j-1]);

        }
    }

    for(int i=1;i<=t;i++)
    {
        if(dp[t][i]>=ans)
            ans=dp[t][i];
    }
    cout<<ans<<endl;

}

再贴一个大神的代码,只用了20行!!! 膜拜

#include<stdio.h>
int main(){
    int n, i, j;
    scanf("%d", &n);
    int **a = new int*[n];
    for(i = 0; i < n; ++i){
        a[i] = new int[i + 1];
        for(j = 0; j <= i; ++j){
            scanf("%d", &a[i][j]);
        }
    }
    for(i = n - 2; i >=0; --i){
        for(j = 0; j <= i; ++j){
            a[i][j] += a[i + 1][j] > a[i + 1][j + 1] ? a[i + 1][j] : a[i + 1][j + 1];
        }
    }
    printf("%d", a[0][0]);
    return 0;
}