POJ 3176 Cow Bowling(DP 数字三角形)

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

思路：这题就是通过经典的从下到上的思想分析，直到推到a[0][0]为止，所以每一步都要选择当前行的当前数与其下一行较大的两个数相加，不断的往上推即可。

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#define LL long long#define inf 0x3f3f3f3fusing namespace std;int a[500][500];int main(){    int n,m,x,k,i,j;    int cla;    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        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++)            {                if(a[i][j]>a[i][j+1])                {                    a[i-1][j]+=a[i][j];                }                else                    a[i-1][j]+=a[i][j+1];            }        }        printf("%d\n",a[0][0]);    }    return 0;}