网格反射 codeforces 249D

看原理的话  到这位大哥的博客  深有体会。。

http://blog.youkuaiyun.com/haha593572013/article/details/8770423

如果碰到网格边界的次数大于n+m-2,那么就能够形成一个完整的棋盘。

View Code
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstdlib>
  4 #include<cstring>
  5 #include<queue>
  6 using std::queue;
  7 using std::min;
  8 int const N = 102000;
  9 int n,m,vis[N][4][4],Re[N][4],Count;
 10 __int64 step;
 11 struct node
 12 {
 13        int x,y,dir;
 14 }cur;
 15 queue<node> Q;
 16 char tmp[5];
 17 int dx[]={-1,-1,1,1};
 18 int dy[]={-1,1,1,-1};
 19 int main()
 20 {
 21     while(~scanf("%d %d",&n,&m))
 22     {
 23           scanf("%d %d",&cur.x,&cur.y);
 24           scanf("%s",tmp);
 25           while(!Q.empty())Q.pop();
 26           if(tmp[0]=='U')
 27              if(tmp[1]=='R')cur.dir=1;
 28              else cur.dir=0;
 29           if(tmp[0]=='D')
 30              if(tmp[1]=='R')cur.dir=2;
 31              else cur.dir=3;
 32 
 33           Q.push(cur);
 34 
 35           memset(vis,0,sizeof(vis));
 36           memset(Re,0,sizeof(Re));
 37           Count=0;
 38 
 39           if(cur.x==1)vis[cur.y][0][cur.dir]=1,Re[cur.y][0]=1,Count=1;
 40           else if(cur.x==n)vis[cur.y][1][cur.dir]=1,Re[cur.y][1]=1,Count=1;
 41 
 42           if(cur.y==1)vis[cur.x][2][cur.dir]=1,Re[cur.x][2]=1,Count=1;
 43           else if(cur.y==m)vis[cur.x][3][cur.dir]=1,Re[cur.x][3]=1,Count=1;
 44 
 45           step=1;
 46           int flag=0;
 47 
 48           while(!Q.empty())
 49           {
 50                 cur=Q.front();
 51                 Q.pop();
 52                 if(cur.dir==0)
 53                 {
 54                    int ff=min(cur.x-1,cur.y-1);
 55                    if(ff==0)
 56                    {
 57                       if(cur.x==1&&cur.y==1)cur.dir=2;
 58                       else
 59                         if(cur.x==1)cur.dir=3;
 60                         else cur.dir=1;
 61                       Q.push(cur);
 62                       continue;
 63                    }
 64                    cur.x-=ff;
 65                    cur.y-=ff;
 66                    step+=ff;
 67                    int inc=0;
 68                    if(cur.x==1)
 69                    {
 70                        if(vis[cur.y][0][cur.dir])flag=1;
 71                        else vis[cur.y][0][cur.dir]=1;
 72                        if(!Re[cur.y][0])Re[cur.y][0]=1,inc=1;
 73                    }
 74                    if(cur.y==1)
 75                    {
 76                        if(vis[cur.x][2][cur.dir])flag=1;
 77                        else vis[cur.x][2][cur.dir]=1;
 78                        if(!Re[cur.x][2])Re[cur.x][2]=1,inc=1;
 79                    }
 80                    if(cur.x==1&&cur.y==1)cur.dir=2;
 81                    else
 82                      if(cur.x==1)cur.dir=3;
 83                      else cur.dir=1;
 84                    if(!flag)Q.push(cur),Count+=inc;
 85                 }
 86                 else
 87                   if(cur.dir==1)
 88                   {
 89                      int ff=min(cur.x-1,m-cur.y);
 90                      if(ff==0)
 91                      {
 92                         if(cur.x==1&&cur.y==m)cur.dir=3;
 93                         else
 94                           if(cur.x==1)cur.dir=2;
 95                           else cur.dir=0;
 96                         Q.push(cur);
 97                         continue;
 98                      }
 99                      step+=ff;
100                      cur.x-=ff;
101                      cur.y+=ff;
102                      int inc=0;
103                      if(cur.x==1)
104                      {
105                          if(vis[cur.y][0][cur.dir])flag=1;
106                          else vis[cur.y][0][cur.dir]=1;
107                          if(!Re[cur.y][0])Re[cur.y][0]=1,inc=1;
108                      }
109                      if(cur.y==m)
110                      {
111                          if(vis[cur.x][3][cur.dir])flag=1;
112                          else vis[cur.x][3][cur.dir]=1;
113                          if(!Re[cur.x][3])Re[cur.x][3]=1,inc=1;
114                      }
115                      if(cur.x==1&&cur.y==m)cur.dir=3;
116                      else
117                        if(cur.x==1)cur.dir=2;
118                        else cur.dir=0;
119                      if(!flag)Q.push(cur),Count+=inc;
120                   }
121                   else
122                     if(cur.dir==2)
123                     {
124                        int ff=min(n-cur.x,m-cur.y);
125                        if(ff==0)
126                        {
127                           if(cur.x==n&&cur.y==m)cur.dir=0;
128                           else
129                             if(cur.x==n)cur.dir=1;
130                             else cur.dir=3;
131                           Q.push(cur);
132                           continue;
133                        }
134                        step+=ff;
135                        cur.x+=ff;
136                        cur.y+=ff;
137                        int inc=0;
138                        if(cur.x==n)
139                        {
140                           if(vis[cur.y][1][cur.dir])flag=1;
141                           else vis[cur.y][1][cur.dir]=1;
142                           if(!Re[cur.y][1])Re[cur.y][1]=1,inc=1;
143                        }
144                        if(cur.y==m)
145                        {
146                           if(vis[cur.x][3][cur.dir])flag=1;
147                           else vis[cur.x][3][cur.dir]=1;
148                           if(!Re[cur.x][3])Re[cur.x][3]=1,inc=1;
149                        }
150                        if(cur.x==n&&cur.y==m)cur.dir=0;
151                        else
152                          if(cur.x==n)cur.dir=1;
153                          else cur.dir=3;
154                        if(!flag)Q.push(cur),Count+=inc;
155                     }
156                     else
157                       if(cur.dir==3)
158                       {
159                          int ff=min(n-cur.x,cur.y-1);
160                          if(ff==0)
161                          {
162                             if(cur.x==n&&cur.y==1)cur.dir=1;
163                             else
164                               if(cur.x==n)cur.dir=0;
165                               else cur.dir=2;
166                             Q.push(cur);
167                             continue;
168                          }
169                          step+=ff;
170                          cur.x+=ff;
171                          cur.y-=ff;
172                          int inc=0;
173                          if(cur.x==n)
174                          {
175                             if(vis[cur.y][1][cur.dir])flag=1;
176                             else vis[cur.y][1][cur.dir]=1;
177                             if(!Re[cur.y][1])Re[cur.y][1]=1,inc=1;
178                          }
179                          if(cur.y==1)
180                          {
181                             if(vis[cur.x][2][cur.dir])flag=1;
182                             else vis[cur.x][2][cur.dir]=1;
183                             if(!Re[cur.x][2])Re[cur.x][2]=1,inc=1;
184                          }
185                          if(cur.x==n&&cur.y==1)cur.dir=1;
186                          else
187                            if(cur.x==n)cur.dir=0;
188                            else cur.dir=2;
189                          if(!flag)Q.push(cur),Count+=inc;
190                       }
191                 if(Count>=n+m-2)break;
192                 if(flag)break;
193           }
194 
195           if(flag)printf("-1\n");
196           else printf("%I64d\n",step);
197     }
198     return 0;
199 }

 

转载于:https://www.cnblogs.com/nuoyan2010/archive/2013/04/08/3008723.html

python+opencv简谱识别音频生成系统源码含GUI界面+详细运行教程+数据 一、项目简介 提取简谱中的音乐信息,依据识别到的信息生成midi文件。 Extract music information from musical scores and generate a midi file according to it. 二、项目运行环境 python=3.11.1 第三方库依赖 opencv-python=4.7.0.68 numpy=1.24.1 可以使用命令 pip install -r requirements.txt 来安装所需的第三方库。 三、项目运行步骤 3.1 命令行运行 运行main.py。 输入简谱路径:支持图片或文件夹,相对路径或绝对路径都可以。 输入简谱主音:它通常在第一页的左上角“1=”之后。 输入简谱速度:即每分钟拍数,同在左上角。 选择是否输出程序中间提示信息:请输入Y或N(不区分大小写,下同)。 选择匹配精度:请输入L或M或H,对应低/中/高精度,一般而言输入L即可。 选择使用的线程数:一般与CPU核数相同即可。虽然python的线程不是真正的多线程,但仍能起到加速作用。 估算字符上下间距:这与简谱中符号的密集程度有关,一般来说纵向符号越稀疏,这个值需要设置得越大,范围通常在1.0-2.5。 二值化算法:使用全局阈值则跳过该选项即可,或者也可输入OTSU、采用大津二值化算法。 设置全局阈值:如果上面选择全局阈值则需要手动设置全局阈值,对于.\test.txt中所提样例,使用全局阈值并在后面设置为160即可。 手动调整中间结果:若输入Y/y,则在识别简谱后会暂停代码,并生成一份txt文件,在其中展示识别结果,此时用户可以通过修改这份txt文件来更正识别结果。 如果选择文件夹的话,还可以选择所选文件夹中不需要识别的文件以排除干扰
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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