poj1163数字三角形之动态规划

最新推荐文章于 2025-03-13 16:01:18 发布

菜涛涛

最新推荐文章于 2025-03-13 16:01:18 发布

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本文链接：https://blog.youkuaiyun.com/qq_44975468/article/details/99565594

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本文探讨了通过动态规划解决数字三角形问题的方法，旨在找到从顶点到底部的最大路径和。介绍了两种实现策略，一种是从顶部到底部递推，另一种是从底部到顶部逆向计算，最终输出最大路径和。

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poj1163数字三角形之动态规划

(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.

InputYour 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. OutputYour 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：给出一个数，表示几行，每i行有i个数字，从上到下，每行只能走相邻的两个数，求从上到下的最大值
思路：明显不是贪心，求全局最优解，用dp。
当时想的是直接按照从上到下做，不确定最后停在哪，需要再求一边最大值

#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std; 
int map[101][101],b[101][101];

int main()
{
    memset(map,0,sizeof(map));
    memset(b,0,sizeof(b));
    int i,j,n;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=i;j++)
        {
            scanf("%d",&map[i][j]);
        }
    }
    b[1][1]=map[1][1];
    int ans=0;
    for(i=2;i<=n;i++)
    {
        for(j=1;j<=i;j++)
        {
            b[i][j]=max(b[i-1][j],b[i-1][j-1])+map[i][j];
            ans=max(b[i][j],ans);
        }
    }
    printf("%d\n",ans);
    return 0;
}

后来想一想其实可以从后往前算，最后一定停在b[1][1]上，直接输出就好了

#include <stdio.h>
#include <stdlib.h> 
#include<string.h>
#include<algorithm>
using namespace std;
int map[101][101],b[101][101];
int main()
{
    memset(map,0,sizeof(map));
    memset(b,0,sizeof(b));
    int i,j,n;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=i;j++)
        {
            scanf("%d",&map[i][j]);
        }
    }
   for(int i=1;i<=n;i++)
   {
            b[n][i]=map[n][i];
   }
    for(i=n-1;i>=1;i--)
    {
        for(j=1;j<=i;j++)
        {
            b[i][j]=max(b[i+1][j],b[i+1][j+1])+map[i][j];
            
        }
    }
    printf("%d\n",b[1][1]);
    return 0;
}