poj1753 Flip Game dfs 枚举 待补完 位运算
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Flip Game
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 41219 Accepted: 17896
Description
Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:
Consider the following position as an example:
bwbw
wwww
bbwb
bwwb
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:
bwbw
bwww
wwwb
wwwb
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.
- Choose any one of the 16 pieces.
- Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).
Consider the following position as an example:
bwbw
wwww
bbwb
bwwb
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:
bwbw
bwww
wwwb
wwwb
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.
Input
The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.
Output
Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).
Sample Input
bwwbbbwbbwwbbwww
Sample Output
4
Source
Northeastern Europe 2000
无限wa流。。。
这题出的真好啊。。。。。
每次感觉自己明白dfs的时候都会遇到一题打脸题///
先贴一下错误的dfs过程:
void dfs(int r,int c,int ans){ if(ans == tot){ if(judge()){ flag = true; } return ; } if(r>4||c>4){ return; } dfs(r+1, c, ans); dfs(r, c+1, ans); flip(r, c); dfs(r+1, c, ans+1); dfs(r, c+1, ans+1); flip(r, c); }
找到了一个反例:
bbbb
bwwb
bwwb
bbbb
用这样的错误代码跑会输出impossible,为啥捏,我原先想象的很形象。。。一个格子是起点呗。。。翻不翻,然后想右走或者向下走。。。
为啥是错的呢,其实脑子里跑一遍这个过程就明白了,先一口气走到第一列最后一排,然后走不了了返回,接着在第三排走到了最后一列。。。。。反正并不能每个格子都check!!
所以还是自己傻啊啊啊啊啊
关于这题,首先要确定下来能在16个操作之内能能解决。每个格子(当然也包括它周围的格子)呢,它翻奇数次就相当于翻一次,而翻偶数次则相当于不翻,因此只要确定这个格子到底翻不翻就行了
所以枚举翻格子次数进行爆搜。。。
搜。。。。
代码:
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <sstream>#include <queue>using namespace std;const int maxn = 10;bool board[maxn][maxn];int dx[4] = {1,-1,0,0};int dy[4] = {0,0,1,-1};bool judge(){ int i,j; for(i=1;i<=4;i++){ for(j=1;j<=4;j++){ if(board[i][j]!=board[1][1]){ return false; } } } return true;}int tot;void flip(int r,int c){ int i; board[r][c] = !board[r][c]; for(i=0;i<4;i++){ board[r+dx[i]][c+dy[i]] = !board[r+dx[i]][c+dy[i]]; }}void print(int ans){ int i,j; cout<<"ans: "<<ans<<endl; for(i=1;i<=4;i++){ for(j=1;j<=4;j++){ cout<<board[i][j]; } cout<<endl; } cout<<"\n\n";}bool flag;void dfs(int r,int c,int ans){ //cout<<r<<" "<<c<<" "<<board[r][c]<<endl; //cout<<"a"; //print(ans); if(ans == tot){ if(judge()){ flag = true; } return ; } if(r>4||flag){ return; } flip(r, c); if(c<4){ dfs(r, c+1, ans+1); }else{ dfs(r+1, 1, ans+1); } flip(r, c); if(c<4){ dfs(r, c+1, ans); }else{ dfs(r+1, 1, ans); } return;}int main(){ int i,j; char unit; for(i=1;i<=4;i++){ for(j=1;j<=4;j++){ cin>>unit; if(unit=='b'){ board[i][j] = true; }else{ board[i][j] = false; } } } flag = false; for(i=0;i<=16;i++){ tot = i; dfs(1, 1, 0); if(flag){ break; } } if(!flag){ cout<<"Impossible\n"; }else{ cout<<i<<endl; } return 0;}
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