CF 812B Sagheer, the Hausmeister

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这篇博客介绍了如何利用动态规划方法解决编程竞赛中的一道题目——CF 812B Sagheer, the Hausmeister。不同于传统的贪婪算法，作者在解题过程中进行了预处理步骤，通过建立简单的方程来确定到达下一层的三种方式，并找出需要到达的最高楼层。" 80250261,7676897,Android广播机制详解,"['Android开发', 'BroadcastReceiver', '系统广播', '编程实践']

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The way I get this done is dynamic programming . Unlike the greedy method I do a little preliminary stuff to finished it.
the equation is pretty easy , just three way of get next floor.but we have to find the topmost floor that need to go .

#include <bits/stdc++.h>
#include <cstring>
using namespace std;
const int maxn = 1005;
const int INF  = 1<<29;
int n = 0,m;
int dp[maxn][2];
int lmost[maxn],rmost[maxn];
int go(int from,int to,int i)
{
    if(from == 1 && to == 1)
    {
        return lmost[i]*2 + 1;
    }
    if(from == 0 && to == 0)
    {
        return rmost[i]*2 + 1;
    }
    if( from != to)
    {
        return m +2;
    }
}
int main()
{
    cin>>n>>m;
    string str[20];
    int ceilling = 0;
    for(int i=n; i>=1; i--)
    {
        cin>>str[i];
    }
    int flag = 0;
    for(int i=n;i>=1;i--)
    {
        for(int j=1;j<=m;j++)
        {
            if(str[i][j] == '1')
            {
                flag = 1;
                break;
            }
        }
        if(flag == 0) n--;
        if(flag) break;
    }
//    cout<<n<<endl;
    for(int i = 1;i<=n;i++)
    {
        int f = 0;
        for(int j=1;j<=m;j++)
        {
            if(str[i][j] == '1')
            {
                f = 1;
                lmost[i] = m-j+1;
                break;
            }
        }
        for(int j = m;j>=1;j--)
        {
            if(str[i][j] == '1')
            {
                rmost[i] = j;
                break;
            }
        }
        if(!f)
        {
            lmost[i] = rmost[i] = 0;
        }
    }
    //cout<<n<<endl;
    if(n == 1)
    {
        cout<<rmost[1]<<endl;
        return 0;
    }
   //cout<<n<<endl;
//    for(int i=n;i>=1;i--)
//    {
//        cout<<lmost[i]<<" "<<rmost[i]<<endl;
//    }
    dp[1][0] = 0;
    dp[1][1] = m;
    for(int i=2; i<=n; i++)
    {
        dp[i][0] = min(dp[i-1][0] + go(0,0,i-1),dp[i-1][1]+go(1,0,i-1));
        dp[i][1] = min(dp[i-1][1] + go(1,1,i-1),dp[i-1][0]+go(0,1,i-1));
    }
//    cout<<endl;
//    for(int i=n;i>=1;i--)
//    {
//        for(int j =0 ;j<2;j++)
//        {
//            cout<<dp[i][j]<<" ";
//        }
//        cout<<endl;
//    }


    cout<<min(dp[n][0] + rmost[n],dp[n][1] + lmost[n])<<endl;


    return 0;
}