描述
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
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输入
- Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99. 输出
- Your program is to write to standard output. The highest sum is written as an integer. 样例输入
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5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
样例输出
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30
经典题目，直接说思路吧。动态规划，从下面往上面处理，题目要求从第一层出最后一层某一个位置的路径的最大和，那么我们求从下到上的最大和。dp[i][j]表示从下面到该位置的最大和，于是状态转移方程：dp[i][j] = max(dp[i-1][j],dp[i-1][j-1])+dp[i][j];
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
int map[101][101];
int dp[101][101];
int n;
using namespace std;
int main(){scanf("%d",&n);memset(dp,0,sizeof(dp));for(int i=1 ;i<=n ;i++){for(int j=1 ;j<=i ;j++){scanf("%d",&map[i][j]);dp[i][j] = map[i][j]; }}for(int i=n-1 ;i>=1 ;i--){//直接从倒数第二层处理的，因为最后一层的下一层都是0 for(int j=n ;j>=1 ;j--){dp[i][j] = max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];}}printf("%d\n",dp[1][1]);return 0;
}