（POJ3176） Cow Bowling <简单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   02   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

题意：
从顶点走到最底层，每次可以走下一层的相邻两个位置，问走的位置上数的和的最大值？

分析：
简单题：
dp[i][j] 表示从(i,j)走到最底层的最大值
从下往上推
dp[i][j] = max（dp[i+1][j]+a[i][j],dp[i+1][j+1]+a[i][j]）；
AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int a[360][360];int dp[360][360];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=i;j++)                scanf("%d",&a[i][j]);        }        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)            dp[n][i] = a[n][i];        for(int i=n-1;i>=1;i--)        {            for(int j=1;j<=i;j++)                dp[i][j] = max(dp[i+1][j]+a[i][j],dp[i+1][j+1]+a[i][j]);        }        printf("%d\n",dp[1][1]);    }    return 0;}