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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:
73 88 1 02 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:
The highest score is achievable by traversing the cows as shown here.
7*3 8*8 1 0*2 7 4 4*4 5 2 6 5
题目大意：
从第一行出发，向下方临近的两个位置行走，直到到达最后一行，得分即路径上各个位置处的分数之和，求最大分数。
解题思路：
从底部向上使用动态规划
dp[k][j] = max(dp[k + 1][j] + map[k][j], dp[k + 1][j + 1] + map[k][j]);
AC代码
#include<iostream>
#include<algorithm>
#include<math.h>
#include<stdio.h>
using namespace std;
int dp[351][351];
int map[351][351];
int main() {int n;while (cin >> n) {for (int i = 1; i <= n; i++) {for (int j = 1; j <= i; j++) {scanf("%d", &map[i][j]);if (i == n) { dp[i][j] = map[i][j]; }else dp[i][j] = 0;}}for (int k = n - 1; k >= 1; k--) {for (int j = 1; j <= k; j++) {dp[k][j] = max(dp[k + 1][j] + map[k][j], dp[k + 1][j + 1] + map[k][j]);}}cout << dp[1][1] << endl;}
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