动态规划(DP)-01背包问题-(百练)1163:The Triangle

描述

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

先用暴力枚举方法:

#include
#include
#include
using namespace std;
int N;
vector<vector<int>>arr;
int f(int i,int j){
	if(i==N)//临界条件 
		return 0;
	return max(f(i+1,j),f(i+1,j+1))+arr[i][j];//要么往下一层的左边 往下一层的右边 
}
int main(){
	cin>>N;
	arr.resize(N);
	for(int i=0;i<N;i++){
		arr[i].resize(N);
		fill(arr[i].begin(),arr[i].end(),0);
	}
	for(int i=0;i<N;i++)
		for(int j=0;j<=i;j++)//输入数据类似于输出直角三角形模板 
		cin>>arr[i][j];
	cout<<f(0,0);
	return 0;
}

进行打表:

这种方法必定 Time Limit Exceeded，但是推动态规划的必要途径,从中可以推出

DPCode:

#include
#include
#include
using namespace std;
int N;
vector<vector<int>>arr;
vector<int>dp;
int main(){
	cin>>N;
	arr.resize(N);
	dp.resize(N+1);
	for(int i=0;i<N;i++){
		arr[i].resize(N);
		fill(arr[i].begin(),arr[i].end(),0);
	}
	for(int i=0;i<N;i++)
		for(int j=0;j<=i;j++)//输入数据类似于输出直角三角形模板 
		cin>>arr[i][j];
	for(int i=N-1;i>=0;--i)
		for(int j=0;j<=i;j++)
			dp[j]=max(dp[j],dp[j+1])+arr[i][j];
	cout<<dp[0];
	return 0;
}