题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5319
Painter
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 826 Accepted Submission(s):
383
Problem Description
Mr. Hdu is an painter, as we all know, painters need
ideas to innovate , one day, he got stuck in rut and the ideas dry up, he took
out a drawing board and began to draw casually. Imagine the board is a
rectangle, consists of several square grids. He drew diagonally, so there are
two kinds of draws, one is like ‘’ , the other is like ‘/’. In each draw he
choose arbitrary number of grids to draw. He always drew the first kind in red
color, and drew the other kind in blue color, when a grid is drew by both red
and blue, it becomes green. A grid will never be drew by the same color more
than one time. Now give you the ultimate state of the board, can you calculate
the minimum time of draws to reach this state.
Input
The first line is an integer T describe the number of
test cases.
Each test case begins with an integer number n describe the number of rows of the drawing board.
Then n lines of string consist of ‘R’ ‘B’ ‘G’ and ‘.’ of the same length. ‘.’ means the grid has not been drawn.
1<=n<=50
The number of column of the rectangle is also less than 50.
Output
Output an integer as described in the problem description.
Each test case begins with an integer number n describe the number of rows of the drawing board.
Then n lines of string consist of ‘R’ ‘B’ ‘G’ and ‘.’ of the same length. ‘.’ means the grid has not been drawn.
1<=n<=50
The number of column of the rectangle is also less than 50.
Output
Output an integer as described in the problem description.
Output
Output an integer as described in the problem
description.
Sample Input
2
4
RR.B
.RG.
.BRR
B..R
4
RRBB
RGGB
BGGR
BBRR
Sample Output
3
6
Source
题目大意：给一个空的长方形，长方形中有很多的正方形方格，现在要给方格涂色，输出最少步数到达，题目输出中给出的最终状态。
特别注意：1、""这个方向的是涂红色。"/"这个方向的是涂蓝色。
2、习惯先涂红色，再涂蓝色。红色和蓝色加起来变成绿色。
3、中间会出现断点的情况，如果出现"."那么就算两步。
详见代码。（选择了一种最直白的方法，没有优化的代码~）
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 int Map[110][110]; 8 char ch[110]; 9 10 int main() 11 { 12 int t; 13 scanf("%d",&t); 14 while (t--) 15 { 16 int n,m; 17 scanf("%d",&n); 18 memset(Map,0,sizeof(Map)); 19 for (int i=0;i<n;i++) 20 { 21 scanf("%s",ch); 22 m=strlen(ch); 23 for (int j=0;j<m;j++) 24 { 25 if (ch[j]=='.') 26 Map[i][j]=0; 27 else if (ch[j]=='R') 28 Map[i][j]=1; 29 else if (ch[j]=='B') 30 Map[i][j]=-1; 31 else 32 Map[i][j]=2; 33 } 34 } 35 int sum=0; 36 for (int i=0;i<n;i++) 37 { 38 for (int j=0;j<m;j++) 39 { 40 int x=i+1,y=j+1; 41 if (Map[i][j]==1) 42 { 43 sum++; 44 while (x<n&&y<m&&Map[x][y]!=0&&Map[x][y]!=-1) 45 { 46 if(Map[x][y]==2) 47 Map[x][y]=-1; 48 else 49 Map[x][y]=0; 50 x++; 51 y++; 52 } 53 } 54 else if (Map[i][j]==-1) 55 { 56 sum++; 57 x=i+1,y=j-1; 58 while (x>=0&&x<n&&y>=0&&y<m&&Map[x][y]!=0&&Map[x][y]!=1) 59 { 60 if(Map[x][y]==2) 61 Map[x][y]=1; 62 else 63 Map[x][y]=0; 64 x++; 65 y--; 66 } 67 } 68 else if (Map[i][j]==2) 69 { 70 sum++; 71 sum++; 72 x=i+1,y=j-1; 73 while (x<n&&y<m&&Map[x][y]!=0&&Map[x][y]!=1) 74 { 75 if(Map[x][y]==2) 76 Map[x][y]=1; 77 else 78 Map[x][y]=0; 79 x++; 80 y--; 81 } 82 x=i+1,y=j+1; 83 while (x>=0&&x<n&&y>=0&&y<m&&Map[x][y]!=0&&Map[x][y]!=-1) 84 { 85 if(Map[x][y]==2) 86 Map[x][y]=-1; 87 else 88 Map[x][y]=0; 89 x++; 90 y++; 91 } 92 } 93 } 94 } 95 printf ("%d ",sum); 96 } 97 return 0; 98 }