POJ 1163 The Triangle + HDU 2084 数塔（递推dp）

最新推荐文章于 2020-03-08 05:49:59 发布

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最新推荐文章于 2020-03-08 05:49:59 发布

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分类专栏：基础线性dp（背包，序列问题)

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本文介绍了一种使用动态规划解决数塔问题的方法，通过从底部向上计算每一步的最大值来找到从顶部到底部的最大路径和。给出了两段C++代码示例，分别针对不同的输入格式。

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B - The Triangle

POJ - 1163

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

Sample Output

简单的动态规划，有递归的思想，从后往前算并记录每一个点的最大值，计算到最后时就是最大的加和

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
int maxy[105][105];
int num[105][105];
int main()
{
    int n;
    cin>>n;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= i; j ++)
        cin>>num[i][j];
    for(int i = 1; i <= n; i ++)
        maxy[n][i] = num[n][i];//给最后一行赋值
    for(int i = n - 1; i >= 1; i --)//从最后一行往回算，计算每一个点的最大值
        for(int j = 1; j <= i; j ++)
        maxy[i][j] = max(maxy[i + 1][j], maxy[i + 1][j + 1]) + num[i][j];//找出并记录每一个点的最大值
    cout<<maxy[1][1]<<endl;//回溯的终点
    return 0;
}

数塔和上面那道题一样的意思

HDU - 2084

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;

int main()
{
    //int dp[105][105];
    int mapp[105][105];
    int t;
    scanf("%d",&t);
    while(t --)
    {
        int n;
        scanf("%d",&n);
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= i; j ++)
            scanf("%d",&mapp[i][j]);
        for(int i = n - 1; i >= 1; i --)
            for(int j = 1; j <= i; j ++)
            {
                mapp[i][j] += max(mapp[i+1][j], mapp[i+1][j+1]);
            }
        printf("%d\n",mapp[1][1]);
    }
    return 0;
}