NYOJ 18 The Triangle 填表法,普通dp
题目链接:
http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=18
The Triangle
时间限制:1000 ms | 内存限制:65535 KB
难度:4
- 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.
- 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
-
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
- 样例输出
-
30
分析:
dp[i][j]=f_max(dp[i+1][j],dp[i+1][j+1])+a[i][j];
代码如下:#include<bits/stdc++.h> #define pai 3.1415926535898 using namespace std; int f_max(int a,int b) { if(a>b) { return a; }else { return b; } } int main() { int n; scanf("%d",&n); int a[n][n]; memset(a,0,sizeof(a)); for(int i=0;i<n;i++) { for(int j=0;j<=i;j++) { scanf("%d",&a[i][j]); } } int dp[n][n]; memset(dp,0,sizeof(dp)); for(int j=0;j<n;j++) { dp[n-1][j]=a[n-1][j]; } for(int i=n-2;i>=0;i--) { for(int j=0;j<=i;j++) { dp[i][j]=f_max(dp[i+1][j],dp[i+1][j+1])+a[i][j]; } } printf("%d\n",dp[0][0]); return 0; }
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