poj 3176 Cow Bowling

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

Sample Output

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.

Source

USACO 2005 December Bronze

基础动态规划，每个数可以加到他下方相邻的两个数上，如此每个数可以加上上方相邻的两个数，选择大的那个。最后选择最大的。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define MAX 351
#define inf 0x3f3f3f3f
using namespace std;
int n;
int s[MAX][MAX];
int main() {
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++) {
        for(int j = 1;j <= i;j ++) {
            scanf("%d",&s[i][j]);
            s[i][j] += max(s[i - 1][j],s[i - 1][j - 1]);
        }
    }
    int ans = 0;
    for(int i = 1;i <= n;i ++) {
        ans = max(ans,s[n][i]);
    }
    printf("%d",ans);
}