sgu 463 - Walking around Berhattan

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原文链接：http://www.cnblogs.com/111qqz/p/4690440.html

在Berhattan街区中，行人沿着街道和大道行走，每经过一个街区，满意度根据街区美观程度变化。本篇介绍了一个算法，用于计算行人完成特定路径后的总满意度。

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K - Walking around Berhattan

Time Limit:250MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice SGU 463

Description

As you probably know, Berhattan is a district of Berland's largest city and it consists of equal square blocks. There are n block lines in the east-west direction and m block lines in the south-north direction. The map shows Berhattan as a rectangle with n rows and mcolumns, so there are nx m blocks in total.

There are n+1 streets running parallel in the east-west direction (horizontally), and there are m+1 avenues running parallel in the south-north direction (vertically). Streets and avenues split the district into blocks and separate Berhattan from other districts of Berland. Each block in Berhattan is characterized by its beauty b _ij.

A pedestrian can walk only along streets and avenues. When the pedestrian walks along any of four sides of a block, we say he passes the block. Every time the pedestrian passes a block his satisfaction is increased by b _ij. If the pedestrian has already passed the block one or more times his satisfaction is increased only by b _ij/2 rounded down when he passes the block again.

You are given the map of Berhattan with the information about the blocks' beauty and the pedestrian's path along the streets and avenues. The path is given as a string containing letters '

', '

' and '

', where '

' means a 90 degree left turn, '

' means a 90 degree right turn, and '

' means walking one block forward by a street or avenue. Facing the east, the pedestrian starts his path in the north-west corner of Berhattan having zero satisfaction level. His path can cross itself and go along the same streets or avenues several times. Pedestrian's satisfaction is increased every time he moves according to the rules described above.

Your task is to calculate the total satisfaction the pedestrian will get after finishing his route.

Picture of the sample test

Input

The first line of input contains two integers n and m (1 ≤ n, m ≤ 100), where n is a number of block lines in Berhattan running in the east-west direction, and m is a number of block lines in Berhattan running in the south-north direction. The following n lines contain m digits each. The j-th digit of the i-th line represents b _ij (0 ≤ b _ij ≤ 9) — the beauty of the corresponding block. The last line of input contains a path in the format specified above. The path consists of 1 up to 500 characters, inclusively. It is guaranteed that the given path doesn't go outside Berhattan.

Output

Print a single integer to the output — the total pedestrian's satisfaction.

Sample Input

sample input

sample output

3 3
123
456
789
MRMMLM

简单模拟,n,m貌似给反了(两个地方给的不一致 ) 害我wa了两发

  1 /*************************************************************************
  2     > File Name: code/2015summer/#5/K.cpp
  3     > Author: 111qqz
  4     > Email: rkz2013@126.com 
  5     > Created Time: 2015年07月30日 星期四 14时00分56秒
  6  ************************************************************************/
  7 
  8 #include<iostream>
  9 #include<iomanip>
 10 #include<cstdio>
 11 #include<algorithm>
 12 #include<cmath>
 13 #include<cstring>
 14 #include<string>
 15 #include<map>
 16 #include<set>
 17 #include<queue>
 18 #include<vector>
 19 #include<stack>
 20 #define y0 abc111qqz
 21 #define y1 hust111qqz
 22 #define yn hez111qqz
 23 #define j1 cute111qqz
 24 #define tm crazy111qqz
 25 #define lr dying111qqz
 26 using namespace std;
 27 #define REP(i, n) for (int i=0;i<int(n);++i)  
 28 typedef long long LL;
 29 typedef unsigned long long ULL;
 30 const int inf = 0x7fffffff;
 31 const int N=1e2+5;
 32 int b[N][N];
 33 int n,m;
 34 char cmd[505];
 35 bool vis[N][N];
 36 int nx,ny;
 37 int dx[4]={-1,0,1,0};
 38 int dy[4]={0,1,-0,-1};
 39 char ch[N][N];
 40 int main()
 41 {
 42     cin>>n>>m;
 43     nx = 0;
 44     ny = 0;
 45     for ( int i = 0 ; i < n ; i++)
 46     cin>>ch[i];
 47     for ( int i = 0 ; i <n ; i++ )
 48     {
 49     for ( int j = 0 ; j < m ; j++ )
 50     {
 51         b[i+1][j+1]=(int)(ch[i][j]-'0');
 52     }
 53     }
 54     int ans  = 0;
 55     memset(vis,false,sizeof(vis));
 56     int dir = 1;
 57     cin>>cmd;
 58     int len = strlen(cmd);
 59     for ( int i = 0 ;  i < len ; i ++ )
 60     {
 61 //    cout<<"ans:"<<ans<<endl;
 62 //    cout<<"nx:"<<nx<<" ny:"<<ny<<endl;
 63     if (cmd[i]=='L')
 64     {
 65         dir = (dir+3)%4;
 66     }
 67     if (cmd[i]=='R')
 68     {
 69         dir  = (dir+1)%4;
 70     }
 71     if (cmd[i]=='M')
 72     {
 73         if (dir==0)
 74         {
 75         if (vis[nx][ny])
 76         {
 77             ans = ans + b[nx][ny]/2;
 78         }
 79         else
 80         {
 81             ans = ans + b[nx][ny];
 82         }
 83         if (vis[nx][ny+1])
 84         {
 85             ans = ans + b[nx][ny+1]/2;
 86         }
 87         else
 88         {
 89             ans = ans + b[nx][ny+1];
 90         }
 91         vis[nx][ny]=true;
 92         vis[nx][ny+1]=true;
 93         nx = nx +dx[dir];
 94         ny = ny +dy[dir];
 95         }
 96         if (dir==2)
 97         {
 98         nx = nx + dx[dir];
 99         ny = ny + dy[dir];
100         if (vis[nx][ny])
101         {
102             ans = ans + b[nx][ny]/2;
103         }
104         else
105         {
106             ans = ans + b[nx][ny];
107         }
108         if (vis[nx][ny+1])
109         {
110             ans = ans + b[nx][ny+1]/2;
111         }
112         else
113         {
114             ans = ans + b[nx][ny+1];
115         }
116         vis[nx][ny]=true;
117         vis[nx][ny+1]=true;
118         }
119         if (dir==1)
120         {
121         nx = nx + dx[dir];
122         ny = ny + dy[dir];
123         if (vis[nx][ny])
124         {
125             ans = ans + b[nx][ny]/2;
126         }
127         else
128         {
129             ans = ans + b[nx][ny];
130         }
131         if (vis[nx+1][ny])
132         {
133             ans = ans + b[nx+1][ny]/2;
134         }
135         else
136         {
137             ans = ans + b[nx+1][ny];
138         }
139         vis[nx][ny]=true;
140         vis[nx+1][ny]=true;
141         
142         }
143             if (dir==3)
144         {
145         
146         if (vis[nx][ny])
147         {
148             ans = ans + b[nx][ny]/2;
149         }
150         else
151         {
152             ans = ans + b[nx][ny];
153         }
154         if (vis[nx+1][ny])
155         {
156             ans = ans + b[nx+1][ny]/2;
157         }
158         else
159         {
160             ans = ans + b[nx+1][ny];
161         }
162         vis[nx][ny]=true;
163         vis[nx+1][ny]=true;
164         nx = nx +dx[dir];
165         ny = ny +dy[dir];
166         }
167     }
168 
169     }
170     cout<<ans<<endl;
171   
172     return 0;
173 }