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  • poj1163The Triangle(动态规划,记忆化搜索)

    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

    这道题若是从上往下搜索会有2n条路径,复杂度很高,但若是换个思路,从下往上搜索,复杂度为O(n2)。
    直接dp代码:

    include

    include

    using namespace std;
    int main()
    {
    int n;
    cin>>n;
    int dp[110][110]={0};
    int a[110][110]={0};
    for(int i=1;i<=n;i++)
    for(int j=1;j<=i;j++)
    cin>>a[i][j];
    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])+a[i][j];
    }
    }
    cout<<dp[1][1]<<endl;
    return 0;
    }

    如果非要从上往下搜索,可以用递推加记忆化搜索,复杂度O(n^2):

    include

    include

    using namespace std;
    int a[110][110]={0},dp[110][110];
    int n;
    int dfs(int i,int j)
    {
    if(i==n) return a[i][j];
    if(dp[i][j]>=0) return dp[i][j];
    dp[i][j]=max(dfs(i+1,j),dfs(i+1,j+1))+a[i][j];
    return dp[i][j];
    }
    int main()
    {
    cin>>n;
    memset(dp,-1,sizeof(dp));
    for(int i=1;i<=n;i++)
    for(int j=1;j<=i;j++)
    cin>>a[i][j];
    dfs(1,1);
    cout<<dp[1][1]<<endl;
    return 0;
    }


    作者:孙建钊
    出处:http://www.cnblogs.com/sunjianzhao/
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。

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  • 原文地址:https://www.cnblogs.com/sunjianzhao/p/11383626.html
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