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POJ 1163 数塔 — 基础DP

2024-09-11 21:26:23   213人阅读

The Triangle

Time Limit: 1000 MS Memory Limit: 10000 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

int a[105][105];

int dp[105][105];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {

        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }

        dp[1][1]=a[1][1];

        for(int i=2;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                dp[i][j]=a[i][j]+max(dp[i-1][j],dp[i-1][j-1]);
                      ///当i不变时，将j列遍历，依次赋给相应的dp[i][j];之后会在下一次
                      ///遍历时再利用max(dp[i-1][j],dp[i-1][j-1])选取max值，直到找到目标
            }
        }
        int res=-999;
        for(int i=1;i<=n;i++)
        {
            res=max(res,dp[n][i]);
        }
        printf("%d\n",res);

    }

    return 0;
}

 

POJ 1163 数塔 — 基础DP


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