Cow Bowling

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 17271	Accepted: 11531

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

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

Sample Output

30

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

點選開啟題目連結http://poj.org/problem?id=3176

題意：

輸入一個n層的三角形，第i層有i個數，求從第1層到第n層的所有路線中，權值之和最大的路線。

規定：第i層的某個數只能連線走到第i+1層中與它位置相鄰的兩個數中的一個,類似楊輝三角...

分析：資料較小，直接用二維陣列存，簡單的dp；dp[i][j]=maps[i][j]+max(dp[i+1][j],dp[i+1][j+1]);

#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<stdio.h>
#include<queue>
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;

#define N 550

int maps[N][N],dp[N][N];

int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        memset(maps,0,sizeof(maps));
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=i;j++)
            {
                scanf("%d",&maps[i][j]);
            }
        }

        for(i=n;i>=1;i--)
        {
            for(j=1;j<=i;j++)
            {
                dp[i][j] = maps[i][j]+max(dp[i+1][j],dp[i+1][j+1]);
            }
        }
        printf("%d\n",dp[1][1]);
    }
    return 0;
}