动态规划
来源:互联网 发布:如何看端口通不通 编辑:程序博客网 时间:2024/06/01 20:02
The Triangle
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 49826 Accepted: 30089
Description
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
Source
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 49826 Accepted: 30089
Description
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
Source
IOI 1994
#include <bits/stdc++.h>#include<algorithm>using namespace std;#define MAX 101int D[MAX][MAX];int n;int maxSum[MAX][MAX];int MaxSum(int i,int j){if(maxSum[i][j]!=-1)return maxSum[i][j];if(i==n) maxSum[i][j]=D[i][j];else {int x=MaxSum(i+1,j);int y=MaxSum(i+1,j+1);maxSum[i][j]=max(x,y)+D[i][j];} return maxSum[i][j];}int main(){int i,j;cin>>n;for(i=1;i<=n;i++) for(j=1;j<=i;j++){ cin>>D[i][j]; maxSum[i][j]=-1;}cout<<MaxSum(1,1)<<endl;}
阅读全文
0 0
- 动态规划!!!动态规划!!!
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- 动态规划
- python scikit-image
- 前端布局引擎bshow
- [bzoj3790]神奇项链 manacher
- JVM 运行时数据区域
- getBytes等空指针异常
- 动态规划
- 读Qt示例之addressbook(一)
- CSS:标准文档流、浮动、绝对定位—(解决有时候父元素不能自动扩展)
- 邮票问题
- socket通信常用函数
- 欢迎使用CSDN-markdown编辑器
- Golang并发编程——安全传输引用和指针的方法
- 优先队列
- 实现图像的边沿检测算法设计