POJ 3176 Cow Bowling（基础DP）

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

573 88 1 02 7 4 44 5 2 6 5

Sample Output

30

经典的动态规划

定义dp[i][j]为从底端到第i行j列元素所经过的路径上元素和的最大值，所以

dp[i][j] = max(dp[i + 1][j],dp[i + 1][j + 1]) + a[i][j]

代码如下：

#include <cstdio>#include <algorithm>using namespace std;const int maxn = 355;int dp[maxn][maxn],a[maxn][maxn];int main(){int n,i,j,k;while(scanf("%d",&n) != EOF){for(i = 0;i < n;i++){for(j = 0;j <= i;j++){scanf("%d",&a[i][j]);}}for(i = n - 1;i >= 0;i--){for(j = 0;j <= i;j++){dp[i][j] = max(dp[i + 1][j],dp[i + 1][j + 1]) + a[i][j];}}printf("%d\n",dp[0][0]);}return 0; }