DP:The Triangle

最新推荐文章于 2020-10-04 20:41:55 发布

lethic

最新推荐文章于 2020-10-04 20:41:55 发布

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The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 29626		Accepted: 17473

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

Sample Output

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>

using namespace std;

int n;
int num[110][110];
int memo[110][110];
int dp(int i,int j){
    int ret=0;
    if(memo[i][j]!=-1)                //记忆化搜索避免重复计算
        return memo[i][j];
    if(i==1)                          //边界，底层即i = 1的层
        return memo[i][j]=num[i][j];
    if(i<1||j>n-i+1||j<1)             //超出界限
        return memo[i][j]=0;
    return memo[i][j]=max(dp(i-1,j),dp(i-1,j+1))+num[i][j];  //在左孩子和右孩子中选一个较大的加上本节点的权值
}
int main(){
    int i,j;
    while(cin>>n){
       for(i=0;i<110;i++)
           for(j=0;j<110;j++)
               memo[i][j]=-1;
       for(i=n;i>0;i--)
           for(j=1;j<=n-i+1;j++)
               scanf("%d",&num[i][j]);
       dp(n,1);
       cout<<memo[n][1]<<endl;
    }
    return 0;
}