Poj 1163 The Triangle 之解题报告

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42232		Accepted: 25527

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

 

对于这题若按照常规可能做不出来；但是用递推的思想就可以计算出来啦

代码：

 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int maxn = 100+10;
int vis[maxn][maxn];
int Triangle[maxn][maxn];

int max(int a,int b)
{
    return a>b?a:b;
}

int main(void)
{
    int i,j,n;
    while(~scanf("%d",&n))
    {
         for(i=1;i<=n;++i)
             for(j=1;j<=i;++j)
                 scanf("%d",&Triangle[i][j]);
         memset(vis,0,sizeof(vis));
         for(i=1;i<=n;++i)
         {
             for(j=1;j<=i;++j)
             {
                  if(i==1) vis[i][j]=Triangle[i][j];
                  else if(j==1) vis[i][j] = vis[i-1][j]+Triangle[i][j];
                  else if(j==i) vis[i][j] = vis[i-1][j-1]+Triangle[i][j];
                  else vis[i][j] = max(vis[i-1][j]+Triangle[i][j],vis[i-1][j-1]+Triangle[i][j]);
             }
         }
         int ans = 0;
         for(i=1;i<n;++i)
             ans = max(ans,vis[n][i]);
         cout<<ans<<endl;
    }
    
    return 0;
}