poj 1163 The Triangle (记忆化搜索)

http://poj.org/problem?id=1163

The Triangle

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

记忆化搜索
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;
typedef long long LL;

#define N 750
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-5
#define met(a, b) memset (a, b, sizeof (a))

int dp

, val

, n;

int DaTaiyang (int x, int y)
{
if (dp[x][y] != -1) return dp[x][y];

if (x>n) dp[x][y] = val[x][y];

dp[x][y] = max (DaTaiyang (x+1, y), DaTaiyang (x+1, y+1)) + val[x][y];

return dp[x][y];
}

int main ()
{
while (scanf ("%d", &n) != EOF)
{
met (dp, -1);

for (int i=1; i<=n; i++)
for (int j=1; j<=i; j++)
scanf ("%d", &val[i][j]);

printf ("%d\n", DaTaiyang (0, 0));
}
return 0;
}