POJ 3176 Cow Bowling Bowling tower problem DP

Title Link: POJ 3176 Cow Bowling

Cow Bowling Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14044 Accepted: 9310 Description the cows don ' t use actual bowling balls when they go bowling. They 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 one, 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. The cow with the highest score wins that 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 above. Source Usaco 2005 December Bronze

Test instructions

From the first level of the tower to the bottom, but only along the diagonal, to find the maximum number of paths on the.

Analysis:

From the bottom up consideration, the size of the and on the path depends directly on the size of the following two numbers. Thus using the top-up approach, step up, go to the top floor, each time choose the largest and, so the final result is saved on the top floor.

State transition equation: dp[i][j] = max (Dp[i+1][j], dp[i+1][j+1]) +a[i][j];

Code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int max_n = n;
int dp[max_n][max_n];
int N, a[max_n][max_n];
void Solve () {
    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] = A[i][j]+max (Dp[i+1][j], dp[i+1][j+1]) ;
    printf ("%d\n", dp[1][1]);
}
int main () {while
    (~scanf ("%d", &n)} {for
        (int i = 1; I <= N; i++) for 
            (int j = 1; J <= I; j + +) 
  SCANF ("%d", &a[i][j]);
        Solve ();
    }
    return 0;
}