HDU 5319多校 模拟
来源:互联网 发布:2016年火灾数据统计 编辑:程序博客网 时间:2024/05/29 10:16
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
24RR.B.RG..BRRB..R4RRBBRGGBBGGRBBRR
Sample Output
36
扫描的时候把G看成B和R的组合。
遇到R或G的时候,向两个方向扫,同时把G变成B。
遇到B或G的时候,同理。
重复扫两遍
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<map>#include<set>#include<cmath>#define ll long longusing namespace std;char Map[60][60];int vis[60][60];int n,m;int in(int x,int y){ if(x>=0&&x<n&&y>=0&&y<m) return 1; return 0;}int sao(int xx,int yy,int c){ int x,y; if(c==1) { x = xx; y = yy; while(1) { if(in(x+1,y+1)&&!vis[x+1][y+1]) { if(Map[x+1][y+1]=='R') { vis[x+1][y+1] = 1; x++,y++; } else if(Map[x+1][y+1]=='G') { Map[x+1][y+1]='B'; x++,y++; } else break; } else break; } x = xx; y = yy; while(1) { if(in(x-1,y-1)&&!vis[x-1][y-1]) { if(Map[x-1][y-1]=='R') { vis[x-1][y-1] = 1; x--,y--; } else if(Map[x-1][y-1]=='G') { Map[x-1][y-1]= 'B'; x--,y--; } else break; } else break; } } else { x = xx; y = yy; while(1) { if(in(x+1,y-1)&&!vis[x+1][y-1]) { if(Map[x+1][y-1]=='B') { vis[x+1][y-1] = 1; x++,y--; } else if(Map[x+1][y-1]=='G') { Map[x+1][y-1] ='R'; x++,y--; } else break; } else break; } x = xx; y = yy; while(1) { if(in(x-1,y+1)&&!vis[x-1][y+1]) { if(Map[x-1][y+1]=='B') { vis[x-1][y+1] = 1; x--,y++; } else if(Map[x-1][y+1]=='G') { Map[x-1][y+1] ='R'; x--,y++; } else break; } else break; } } return 0;}int main(){ int t; cin>>t; while(t--) { memset(vis,0,sizeof(vis)); scanf("%d",&n); for(int i=0; i<n; i++) scanf("%s",Map[i]); m = strlen(Map[0]); int sum = 0; for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { if(vis[i][j]) continue; if(Map[i][j]=='.') continue; sum++; if(Map[i][j]=='R'||Map[i][j]=='G') { if(Map[i][j]=='R') vis[i][j] = 1; else Map[i][j]='B'; sao(i,j,1); } else if(Map[i][j]=='B'||Map[i][j]=='G') { if(Map[i][j]=='B') vis[i][j] = 1; else Map[i][j]='R'; sao(i,j,0); } } } for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { if(vis[i][j]) continue; if(Map[i][j]=='.') continue; sum++; if(Map[i][j]=='R'||Map[i][j]=='G') { if(Map[i][j]=='R') vis[i][j] = 1; else Map[i][j]='B'; sao(i,j,1); } else if(Map[i][j]=='B'||Map[i][j]=='G') { if(Map[i][j]=='B') vis[i][j] = 1; else Map[i][j]='R'; sao(i,j,0); } } } printf("%d\n",sum); } return 0;}
0 0
- HDU 5319多校 模拟
- HDU 5319 Painter(模拟)
- HDU 5319模拟
- Hdu 5319 Painter (模拟)
- HDU 5319 Painter (模拟 脑洞题)
- 模拟(HDU 5319,Painter)
- HDU 5336多校 十滴水模拟
- HDU 5818 (多校 7) 模拟
- hdu 5319 Painter(模拟)(思维)(模拟)
- hdu 5319 Painter(模拟题)
- HDU 5387(2015多校8)-Clock(模拟)
- HDU 5319 两把刷子刷墙问题(多校)-直接模拟
- 【几何模拟】hdu 3286
- hdu 1020 Encoding(模拟)
- hdu 1303 Doubles(模拟)
- hdu 4020 模拟
- HDU **** 暴力模拟
- HDU 4068 模拟
- WPF中的柱形图的创建与使用
- web socket实现聊天室
- iOS分分钟搞定C语言 —— 进制
- Mobile OpenCart 自适应主题模板 ABC-0074
- NoSQL数据库使用心得
- HDU 5319多校 模拟
- codeforce 550 D Regular Bridge
- Java反射—模拟Spring的Aop
- linux目录结构
- Kafka集群安装
- 多态,重载和覆盖
- 提高sql查询性能-使用instr函数替换like
- iOS面试笔试汇总
- linux初学(十三)之linux权限机制