POJ 3176 (DP Pyramid) _ Dynamic Programming DP

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 as this:

          7



        3   8



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

Then the other cows traverse the triangle starting the ' its tip and moving ' down ' to one of the two diagonally adjacent Co WS 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 this frame.

Given a triangle with n (1 <= n <=) 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.

Source

Click to open the topic link http://poj.org/problem?id=3176

Meaning

Enter an n-tier triangle, the I layer has number I, to find from the 1th layer to the N layer of all the routes, weights and the largest route.

Rule: A number of layer I can only be connected to the I+1 layer and its position adjacent to the two number of one, similar to the Yang Hui's triangle ...

Analysis: Small data, directly using two-dimensional array storage, simple 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 LL;
using namespace std;

#define N

Maps[n][n],dp[n][n int];

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;
}