poj3176

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19801 Accepted: 13123

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

思路：从下往上走，最后使得总和最大，其实就是上一层从下一层选择大的加上去，不断更新最后的出最大的。

代码：

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;int a[355][355];int main(){    int N;    while(scanf("%d",&N)!=EOF)    {        int i,j;        for(i=1;i<=N;i++)        {            for(j=1;j<=i;j++)            {                scanf("%d",&a[i][j]);            }        }        for(i=N-1;i>=1;i--)        {            for(j=1;j<=i;j++)            {                a[i][j]=a[i][j]+max(a[i+1][j+1],a[i+1][j]);            }        }        cout<<a[1][1]<<endl;    }    return 0;}