Light oj 1004 - Monkey Banana Problem(DP)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1004

1004 - Monkey Banana Problem

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You are in the world of mathematics to solve the great "Monkey Banana Problem". It states that, a monkey enters into a diamond shaped two dimensional array and can jump in any of the adjacent cells down from its current position (see figure).
While moving from one cell to another, the monkey eats all the bananas kept in that cell. The monkey enters into the array from the upper part and goes out through the lower part. Find the maximum number of bananas the monkey can eat.

Input

Input starts with an in
4000
teger T (≤ 50), denoting the number of test cases.

Every case starts with an integer N (1 ≤ N ≤ 100). It denotes that, there will be 2*N - 1 rows. The ith (1 ≤ i ≤ N) line of next N lines contains exactly i numbers.
Then there will be N - 1 lines. The jth (1 ≤ j < N) line contains N - j integers. Each number is greater than zero and less than 215.

Output

For each case, print the case number and maximum number of bananas eaten by the monkey.

Sample Input	Output for Sample Input
2 4 7 6 4 2 5 10 9 8 12 2 2 12 7 8 2 10 2 1 2 3 1	Case 1: 63 Case 2: 5

Note

Dataset is huge, use faster I/O methods.

题目大意：求从上到下路径上的最大值

解析：简单的DP

代码如下：

#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<set>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#define N 1009
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef long long LL;
int dp

, a

;

int main()
{
int t, n, i, j, cnt = 0;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
for(i = 1; i <= n; i++)
{
for(j = 1; j <= i; j++)
scanf("%d", &a[i][j]);
}
int k = 2 * n - 1, nn = n;
for(i = n + 1; i <= k; i++)
{
nn--;
for(j = 1; j <= nn; j++)
scanf("%d", &a[i][j]);
}
memset(dp, 0, sizeof(dp));
for(i = 1; i <= n; i++)
{
for(j = 1; j <= i; j++)
{
dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j]) + a[i][j];
}
}
for(i = n + 1; i <= k; i++)
{
for(j = 1; j <= n; j++)
dp[i][j] = max(dp[i - 1][j + 1], dp[i - 1][j]) + a[i][j];
}
printf("Case %d: %d\n", ++cnt, dp[k][1]);
}
return 0;
}