簡單dp算法——Cow Bowling

A - Cow Bowling

點擊打開鏈接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.

解題思路：

       從倒數第二行開始，m[n-1][j] += max( m[n][j]，m[n][j+1] )，依次往上選出最大的數，最後剩下m[1][1]，也就是最大的那個數。

代碼實現：

#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;
}