zoukankan      html  css  js  c++  java
  • ZOJ 3141 Arnie's Dog Biscuits【dp递推】

    Arnie's Dog Biscuits

    Time Limit: 1 Second      Memory Limit: 32768 KB

    Our discerning gourmet puppy Arnie is turning to you for a program to help him split his dog biscuits. Each biscuit is shaped like a rectangle and perforated into equal sized squares:

    Unfortunately, Arnie will only eat square-shaped biscuits; therefore, he must break the biscuit into squares. Each break, termed a split, is applied to one rectangle, runs along one straight perforated line, and separates the rectangle into two pieces:

    Input

    The first line of the input contains one positive integer n, the number of biscuits to split. Each of the next n lines contains two positive integers r and c, the number of rows and columns of one biscuit, separated by white space.

    Output

    The output contains one line for each biscuit specifying the minimal number of splits required to break the biscuit into squares.

    Sample Input

    2
    6 7
    5 5
    

    This defines two biscuits: the one shown above which requires four splits, and a square biscuit which requires no splits.

    Sample Output

    4
    0
    

    dp[i][j]=min(dp[r][j]+dp[i-r][j]+1, dp[i][k]+dp[i][j-k]+1) 

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define ms(x,y) memset(x,y,sizeof(x))
    using namespace std;
    
    typedef long long ll;
    
    const double pi = acos(-1.0);
    const int mod = 1e9 + 7;
    const int maxn = 1e5 + 5;
    
    int dp[1100][1100];
    
    int main()
    {
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
    
        int n, m;
        ms(dp, INF);
        for (int i = 0; i <= 250; i++)
        {
            dp[i][i]=0;
            dp[i][0]=0;
            dp[0][i]=0;
        }
        for (int i = 1; i <= 250; i++)
        {
            for (int j = 1; j <= 250; j++)
            {
                for (int k = 0; k <= i; k++)
                {
                    dp[i][j] = min(dp[i][j], dp[k][j] + dp[i - k][j] + 1);
                }
                for (int k = 0; k <= j; k++)
                {
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[i][j - k] + 1);
                }
            }
        }
    
        int t;
        scanf("%d", &t);
        while (t--)
        {
            int n, m;
            scanf("%d%d", &n, &m);
            printf("%d
    ", dp[n][m]);
        }
        return 0;
    }


    Fighting~
  • 相关阅读:
    跳转指定页面
    如何解决项目中.a文件中的.o冲突
    地图根据起点和终点计算中心点角度来绘制弧线 iOS
    codePush常用
    ios原生push到RN界面后pop
    atomic,nonatomic的区别
    KVC
    jQuery绑定event事件的各种方法比较
    Git常用命令总结
    多个$(document).ready()的执行顺序问题
  • 原文地址:https://www.cnblogs.com/Archger/p/12774693.html
Copyright © 2011-2022 走看看