zoukankan      html  css  js  c++  java
  • UVALive 4425 Another Brick in the Wall 暴力

    C - Another Brick in the Wall
    Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

    Description

    Download as PDF

    After years as a brick-layer, you've been called upon to analyze the instability of brick walls. The instability of a wall can be approximated by the maximum damage to a wall in case of taking one brick out. A brick will fall if all bricks that are directly underneath it are removed. Note that if the space underneath a brick is partially empty, it does not fall. You are given the description of all bricks in a wall, and must determine the instability of the wall as described in the following sections.

    Input

    There are multiple test cases in the input. Each test case consists of a single line, ``M N " (1$ le$M, N$ le$100) where M and N indicate the height and width (in units), respectively, of the input wall.


    Each of the next M lines is a string of N digits which specifies a row in the wall. Each brick in a row is represented by a substring of the row (like s ) such that every digit in s is the same, which is equal to the length of s too. For example, 333 and 22 are two bricks of length 3 and 2 respectively, but 111 specifies three bricks of length one. A 0 in a row means there is no brick in that place of wall. Note that the height of each brick is one. The input terminates with a line containing ` 0 0'. You may assume that the input is correct. This means:

    1. There is no brick such that the length of the brick does not conform to the digits in the brick (like 222 in the row 12221).
    2. No brick can fall initially.

    Output

    For each test case, write a single line containing maximum sum of the bricks' lengths that will fall if we take one brick out (including that brick).

    Sample Input

    4 5 
    33322 
    22333 
    33322 
    22333 
    4 6 
    122333 
    444422 
    111111 
    333333 
    3 3 
    022 
    220 
    111 
    0 0
    

    Sample Output

    5 
    8 
    4
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 100001
    const int inf=0x7fffffff;   //无限大
    int dp[110][110];
    int g[110][110];
    int q[110][110];
    int main()
    {
        int n,m;
        string s[110];
        while(cin>>n>>m)
        {
            if(n==0&&m==0)
                break;
            memset(dp,0,sizeof(dp));
            memset(q,0,sizeof(q));
            memset(g,0,sizeof(g));
            for(int i=0;i<n;i++)
            {
                cin>>s[i];
                for(int j=0;j<m;j++)
                {
                    if(s[i][j]=='0')
                    {
                        g[i][j]=-1;
                    }
                    else
                    {
                        g[i][j]=(int)(s[i][j]-'0');
                        j=j+(int)(s[i][j]-'0')-1;
                    }
                }
            }
    
    
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    if(i==0)
                    {
                        dp[i][j]=(int)(s[i][j]-'0');
                    }
                    else
                    {
                        if(g[i][j]==-1)
                            dp[i][j]=0;
                        else if(g[i][j]==0)
                            dp[i][j]=dp[i][j-1];
                        else
                        {
                            int ii=i-1;
                            int sum=0;
                            int op=g[i][j];
                            dp[i][j]+=g[i][j];
    
                            for(int k=0;k<m;k++)
                            {
                                if(j<=k&&k<j+op)
                                    q[i][k]=-1;
                                else
                                    q[i][k]=g[i][k];
                            }
    
    
                            while(1)
                            {
                                sum=0;
    
                                for(int k=0;k<m;k++)
                                {
                                    if(q[ii][k]==-1)
                                        continue;
                                    q[ii][k]=g[ii][k];
                                    for(int jj=0;jj<g[ii][k];jj++)
                                    {
                                        if(q[ii+1][k+jj]!=-1)
                                            break;
                                        if(jj==g[ii][k]-1)
                                        {
                                            for(int mm=0;mm<g[ii][k];mm++)
                                            {
                                                q[ii][k+mm]=-1;
                                                sum++;
                                            }
                                        }
                                    }
                                }
                                dp[i][j]+=sum;
                                if(sum==0)
                                    break;
                                ii--;
                                if(ii<0)
                                    break;
                            }
                            memset(q,0,sizeof(q));
                        }
                    }
                }
            }
            int ans=0;
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    ans=max(ans,dp[i][j]);
                }
            }
            cout<<ans<<endl;
        }
        return 0;
    }

  • 相关阅读:
    memwatch内存泄露检测工具
    JavasSript实现秒转换为“天时分秒”控件和TDD测试方法应用
    字符编码转换笔记
    AjaxFileUpload 方法与原理分析
    Lua Rings库介绍
    Virtualbox+UbuntuServer+Xshell搭建Linux开发环境
    HTTP下载文件名称编码说明
    lua metatable 和 _index 实验
    前向后瞻正则表达式及其JS例子
    浏览器浏览记忆(history)几则技巧记录
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4238233.html
Copyright © 2011-2022 走看看