简单dp算法——Cow Bowling

A -
Cow Bowling

点击打开链接http://acm.hust.edu.cn/vjudge/contest/123760#problem/A

Crawling in process...
Crawling failed
Time Limit:1000MS
Memory Limit:65536KB
64bit IO Format:%lld & %llu

Submit
Status

Description

Input

Output

Sample Input

Sample Output

Hint

Description

The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this:

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.

Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.

Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.

Output

Line 1: The largest sum achievable using the traversal rules

Sample Input

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

Sample Output

30

Hint

Explanation of the sample:

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.

解题思路：
从倒数第二行开始，m[n-1][j] += max( m
[j]，m
[j+1] )，依次往上选出最大的数，最后剩下m[1][1]，也就是最大的那个数。

代码实现：

#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;

int main()
{
int n;
cin >> n;
long long m[400][400] = { 0 };
int i, j;
for( i=1; i<=n; i++ )
{
for( j=1; j<=i; j++ )
{
cin >> m[i][j];
}
}

for( i=n-1; i>0; i-- )
{
for( j=1; j<=i; j++)
{
m[i][j] += max( m[i+1][j], m[i+1][j+1] );
}
}
cout << m[1][1] <<endl;
return 0;
}