zoukankan      html  css  js  c++  java
  • (记忆化搜索 )The Triangle--hdu --1163

     
     
    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


    单纯的递归, 但是会超时

    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    #define N 110
    #define max(a,b) (a>b?a:b)
    
    int a[N][N];
    
    int DFS(int x, int y, int n)
    {
        if(x>n || y>n)
            return 0;
    
        if(x==n && y==n)
            return a[x][y];
    
        return a[x][y]+ max(DFS(x+1,y, n), DFS(x+1, y+1, n));
    }
    
    int main()
    {
        int n;
    
        while(scanf("%d", &n)!=EOF)
        {
            int i, j;
    
            memset(a, 0, sizeof(a));
    
            for(i=1; i<=n; i++)
            for(j=1; j<=i; j++)
                scanf("%d", &a[i][j]);
    
            printf("%d
    ", DFS(1,1,n));
        }
        return 0;
    }

    用上记忆化搜索后, 不会超时了

    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    #define N 110
    #define max(a,b) (a>b?a:b)
    
    int a[N][N], dp[N][N];
    
    int  DFS(int x, int y, int n)
    {
        if(x>n || y>n)
            return 0;
    
        if(dp[x][y]!=-1)
            return dp[x][y];
        else
        {
            if(x==n && y==n)
            return a[x][y];
    
            dp[x+1][y] = DFS(x+1, y, n);
            dp[x+1][y+1] = DFS(x+1, y+1, n);
    
            return a[x][y]+ max(dp[x+1][y], dp[x+1][y+1]);
        }
    }
    
    int main()
    {
        int n;
    
        while(scanf("%d", &n)!=EOF)
        {
            int i, j;
    
            memset(a, 0, sizeof(a));
            memset(dp, -1, sizeof(dp));
    
            for(i=1; i<=n; i++)
            for(j=1; j<=i; j++)
                scanf("%d", &a[i][j]);
    
            printf("%d
    ", DFS(1, 1, n));
        }
        return 0;
    }



    勿忘初心
  • 相关阅读:
    团队作业 总结
    个人作业 Alpha项目测试
    第二次作业
    交互式多媒体图书平台的设计与实现
    基于VS Code的C++语言的构建调试环境搭建指南
    码农的自我修养之必备技能 学习笔记
    工程化编程实战callback接口学习
    如何测评一个软件工程师的计算机网络知识水平和编程能力
    深入理解TCP协议及其源代码
    Socket与系统调用深度分析
  • 原文地址:https://www.cnblogs.com/YY56/p/5038431.html
Copyright © 2011-2022 走看看