POJ 1163 The Triangle & POJ 3176 Cow Bowling(DP动态规划)

买一送一的动态规划题目，让我开心地AC两次！唯一改变的就是数组的大小。

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

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.

#include<iostream>#include<cmath>using namespace std;const int MAX = 350 + 10;int main(){int t[MAX][MAX];int dp[MAX][MAX];memset(dp,0,sizeof(dp));int n;cin >> n;for(int i = 0; i < n; ++i)for(int j = 0; j < i + 1; ++j){cin >> t[i][j];dp[i][j] = t[i][j];}for(int i = n - 2;i >= 0; --i){int j = i + 1;for(int x = 0; x < j; ++x) dp[i][x] = max(t[i][x] + dp[j][x],t[i][x] + dp[j][x + 1]);}cout << dp[0][0] << endl;return 0;}