zoukankan      html  css  js  c++  java
  • hdu 2125 Local area network 简单dp之pair初体验

    A local area network (LAN) supplies networking capability to a group of computers in close proximity to each other such as in an office building, a school, or a home. A LAN is useful for sharing resources like files, printers, games or other applications. A LAN in turn often connects to other LANs, and to the Internet or other WAN.
    In this contest, everybody is connecting with each others like the following figure.

    The contest’s network was built as N rows and M columns, your computer locate at (0, 0) and the judger’s computer locate at (m-1, n-1) The code you submit would only transfer to up or right smoothly. It doesn’t matter if some accidents happened. Could you tell me how many ways from your computer to judger when a data wire was broken?

    Input
    There are multiple cases. Every case contains two integers in the first line, N, M (3<=N+M<=40). Second line contains four integers X1, Y1, X2, Y2 (0<=X1, X2<M, 0<=Y1, Y2<N, |X1+Y1-X2-Y2|=1), meaning the broken wire position.
    Output
    Print the answer in one line.
    Sample Input
    3 3
    0 0 1 0
    Sample Output
    3

    并不是什么难题,看一眼就知道的简单dp思想
    想找个机会使用一下pair
    然而这道其实没有必要使用
    记一下pair语法
    声明和赋值
    pair<int ,int >p (5,6);
    pair<int ,int > p1= make_pair(5,6);
    pair<string,double> p2 (“aa”,5.0);
    引用时.second
    值得一提的是 pair 比较规则为
    先比较first 再比较second
    上代码

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    #include<iostream>
    using namespace std;
    long long dp[45][45];
    void print(int n,int m)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
                printf("%lld ",dp[i][j]);
            cout<<endl;
        }
        cout<<endl;
        return ;
    }
    int main()
    {
    
        int n,m;
        while(~scanf("%d%d",&n,&m))
        {
            int flag=0;//1 heng 2 shu
            pair<int,int>coor1,coor2,coor3;
            int tmp1,tmp2,tmp3,tmp4;
            scanf("%d%d%d%d",&tmp1,&tmp2,&tmp3,&tmp4);
            if(tmp1==tmp3) flag=1;
            else flag=2;
            coor1=make_pair(tmp1,tmp2);
            coor2=make_pair(tmp3,tmp4);
            coor3=max(coor1,coor2);
            //cout<<coor3.first<<endl;
            //cout<<coor3.second<<endl;
            dp[0][0]=1;
            for(int i=1; i<n; i++)
            {
                if(flag==1&&coor3.second==i&&coor3.first==0)
                    dp[0][i]=0;
                else dp[0][i]=dp[0][i-1];
            }
            for(int i=1; i<m; i++)
            {
                if(flag==2&&coor3.first==i&&coor3.second==0)
                    dp[i][0]=0;
                else dp[i][0]=dp[i-1][0];
            }
            if(m==1||n==1)
            {
                printf("%lld
    ",dp[n-1][m-1]);
                continue;
            }
            for(int i=1; i<m; i++)
            {
                for(int j=1; j<n; j++)
                {
                    if(coor3.first==i&&coor3.second==j)
                    {
                        if(flag==1) dp[i][j]=dp[i-1][j];
                        else dp[i][j]=dp[i][j-1];
                    }
                    else dp[i][j]=dp[i-1][j]+dp[i][j-1];
                }
            }
            //print(n,m);
            printf("%lld
    ",dp[m-1][n-1]);
        }
        return 0;
    }
    //caowenbo
    
  • 相关阅读:
    公告Ext3.3Bate发布了——PivotGrids, Calendars and 更多其他改进
    纪念一起工作三年的朋友的离开
    基于Ext.Panel扩展一个BMap
    Bing Map App 开发 还没入门遇见错误无法继续
    遗传算法学习笔记(2)
    接下来的目标
    Silverlight游戏开发心得(4)——重读调度器
    遗传算法学习笔记(5)
    Silverlight游戏开发心得(3)——有限状体机
    遗传算法学习笔记(4)
  • 原文地址:https://www.cnblogs.com/caowenbo/p/11852339.html
Copyright © 2011-2022 走看看