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
这道题若是从上往下搜索会有2n条路径，复杂度很高，但若是换个思路，从下往上搜索，复杂度为O(n2)。
直接dp代码：
include
include
using namespace std;
int main()
{
int n;
cin>>n;
int dp[110][110]={0};
int a[110][110]={0};
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
cin>>a[i][j];
for(int i=n;i>=1;i--)
{
for(int j=1;j<=i;j++)
{
dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+a[i][j];
}
}
cout<<dp[1][1]<<endl;
return 0;
}
如果非要从上往下搜索，可以用递推加记忆化搜索，复杂度O(n^2)：
include
include
using namespace std;
int a[110][110]={0},dp[110][110];
int n;
int dfs(int i,int j)
{
if(i==n) return a[i][j];
if(dp[i][j]>=0) return dp[i][j];
dp[i][j]=max(dfs(i+1,j),dfs(i+1,j+1))+a[i][j];
return dp[i][j];
}
int main()
{
cin>>n;
memset(dp,-1,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
cin>>a[i][j];
dfs(1,1);
cout<<dp[1][1]<<endl;
return 0;
}