标签：dp
Description
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow‘s score is the sum of the numbers of the cows visited along the way. The cow with the highest
score wins that frame. Input
Output
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
Hint
7
*
3 8
*
8 1 0
*
2 7 4 4
*
4 5 2 6 5
The highest score is achievable by traversing the cows as shown above.Source
记忆化搜索：
/*47ms,1356KB*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 355;
int a[maxn][maxn], dp[maxn][maxn], n;
int f(int i, int j)
{
if (dp[i][j] >= 0) return dp[i][j];
if (i == n - 1) return dp[i][j] = a[i][j];
return dp[i][j] = a[i][j] + max(f(i + 1, j), f(i + 1, j + 1));
}
int main()
{
int i, j;
scanf("%d", &n);
for (i = 0; i < n; ++i)
for (j = 0; j <= i; ++j)
scanf("%d", &a[i][j]);
memset(dp, -1, sizeof(dp));
printf("%d", f(0, 0));
return 0;
}标签：dp
原文地址：http://blog.csdn.net/acm_10000h/article/details/41120631
