zoukankan      html  css  js  c++  java
  • hdu_2391 Filthy Rich DP

    Filthy Rich

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4393    Accepted Submission(s): 1898


    Problem Description
    They say that in Phrygia, the streets are paved with gold. You’re currently on vacation in Phrygia, and to your astonishment you discover that this is to be taken literally: small heaps of gold are distributed throughout the city. On a certain day, the Phrygians even allow all the tourists to collect as much gold as they can in a limited rectangular area. As it happens, this day is tomorrow, and you decide to become filthy rich on this day. All the other tourists decided the same however, so it’s going to get crowded. Thus, you only have one chance to cross the field. What is the best way to do so?

    Given a rectangular map and amounts of gold on every field, determine the maximum amount of gold you can collect when starting in the upper left corner of the map and moving to the adjacent field in the east, south, or south-east in each step, until you end up in the lower right corner.
     
    Input
    The input starts with a line containing a single integer, the number of test cases.
    Each test case starts with a line, containing the two integers r and c, separated by a space (1 <= r, c <= 1000). This line is followed by r rows, each containing c many integers, separated by a space. These integers tell you how much gold is on each field. The amount of gold never negative.
    The maximum amount of gold will always fit in an int.
     
    Output
    For each test case, write a line containing “Scenario #i:”, where i is the number of the test case, followed by a line containing the maximum amount of gold you can collect in this test case. Finish each test case with an empty line.
     
    Sample Input
    1
    3 4
    1 10 8 8
    0 0 1 8
    0 27 0 4
     
    Sample Output
    Scenario #1:
    42

    一道dp题, 从左上角走到右下角,只能往右、下或右下走,求捡到的金子最多为多少。

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    
    using namespace std;
    
    int dp[1200][1200];
    
    int Max(int a, int b)
    {
        return a>b? a:b;
    }
    
    int main()
    {
        int t, r, c;
        scanf("%d", &t);
        for(int i=0; i<1200; i++)
        {
            dp[i][0]=0;
            dp[0][i]=0;
        }
        for(int k=0; k<t; k++)
        {
            scanf("%d%d", &r, &c);
    
            for(int i=1; i<=r; i++)
                for(int j=1; j<=c; j++)
            {
                scanf("%d", &dp[i][j]);
                dp[i][j] += Max(dp[i-1][j], dp[i][j-1]);
            }
            printf("Scenario #%d:
    ", k+1);
            printf("%d
    
    ", dp[r][c]);
        }
    
        return 0;
    }
  • 相关阅读:
    回顾
    单例模式
    元类
    反射和内置方法
    issubclass 和 isinstance和断点调试
    绑定方法和非绑定方法
    并发编程:IO多路复用。
    基于tcp的下载文件,以及struct模块的应用。
    并发编程:协程,异步调用。
    并发编程:GIL,线程池,进程池,阻塞,非阻塞,同步,异步
  • 原文地址:https://www.cnblogs.com/Dawn-bin/p/10802325.html
Copyright © 2011-2022 走看看