POJ-1163-The Triangle
2017-04-07 14:44
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题目:
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.
Input
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.
Output
Your program is to write to standard output. The highest sum is written as an integer.
Sample Input
Sample Output
大概意思是:从数字三角形的顶端开始,寻找一条路径,使这条路径上每个结点数值之和最大,规定在每个节点处只能向左下或右下行进。
解析:这道题当然是用动态规划来做啦!建立一个数组dp[][]用来存储从每个结点开始的最大值路径,然后运用状态转移方程:dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])就可以啦!题目和代码都比较简单。
下面是AC过得代码:
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.
Input
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.
Output
Your program is to write to standard output. The highest sum is written as an integer.
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
大概意思是:从数字三角形的顶端开始,寻找一条路径,使这条路径上每个结点数值之和最大,规定在每个节点处只能向左下或右下行进。
解析:这道题当然是用动态规划来做啦!建立一个数组dp[][]用来存储从每个结点开始的最大值路径,然后运用状态转移方程:dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])就可以啦!题目和代码都比较简单。
下面是AC过得代码:
#include <iostream> #include <cmath> #include <cstring> using namespace std; int main() { int n;//表示三角形有n层 cin>>n; int tri[n+1][n+1]; int dp[n+2][n+2]; for(int i=1;i<=n;i++) for(int j=1;j<=i;j++) cin>>tri[i][j]; memset(dp,0,sizeof(dp)); for(int i=n;i>=1;i--) { for(int j=1;j<=i;j++) dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+tri[i][j]; } cout<<dp[1][1]<<endl; return 0; }
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