6581 Number Triangle

6581 Number Triangle

时间限制:500MS  内存限制:1000K
提交次数:57 通过次数:47

题型: 编程题   语言: G++;GCC

 

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. 

输入格式

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 &gt; 1 but <= 100. 
The numbers in the triangle, all integers, are between 0 and 99. 

输出格式

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

 

输入样例

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

 

输出样例

30

 

作者

 admin

 

　　最原始的数塔问题，，经典的动态规划问题。状态转移方程：dp[i][j]=max(dp[i-1][j-1],dp[i-1][j])+a[i][j];

其他细节见代码注释：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;

int a[105][105];//a[i][i]为数塔上点[i][i]处输入的值
int dp[105][105];//dp[i][j]为走到点[i][j]所能得的最大值
int main()
{
    //freopen("input.txt","r",stdin);
    memset(a,0,sizeof(a));  //初始化数组
    memset(dp,0,sizeof(dp));
    int n;
    scanf("%d",&n);  //输入数据
    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];
    if(n==1) //n为1就直接输入塔顶数字
    {
        printf("%d\n",dp[1][1]);
        return 0;
    }
    //状态转移方程
    for(int i=2;i<=n;i++)
    {
        for(int j=1;j<=i;j++)
        { 
          //走到第a[i][j]位置能得到的最大值dp[i][j]就是等于选择从走到a[i-1][j-1]和走到
          //a[i-1][j]中值较大的那种走法里再加上a[i][j];
            dp[i][j]=max(dp[i-1][j-1],dp[i-1][j])+a[i][j];
        }
    }
    //
    int Max=-1;
    for(int i=1;i<=n;i++)//扫描塔的最下层，找出最大值
        if(dp[n][i]>Max)
            Max=dp[n][i];
    printf("%d\n",Max);
    return 0;
}