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題目鏈接：http://poj.org/problem?id=1753

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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:?

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

bwwb bbwb bwwb bwww

Sample Output

4

題意：找出能使4X4的棋盤中的棋子同色的最小步數。

代碼例如以下：

#include <iostream> #include <cstring> using namespace std; int chess[7][7];//事實上利用的僅僅有中心的4x4 int x[5] = {0,0,1,0,-1}; int y[5] = {0,1,0,-1,0}; int flag, step; int judge(int chess[7][7])//推斷顏色是否所有同樣 {for(int i = 1; i <= 4; i++){for(int j = 1; j <= 4; j++){if(chess[i][j] != chess[1][1])return 0;}}return 1; } void flip(int row, int col)//翻棋 {for(int i = 0; i <= 4; i++){if(chess[row+x[i]][col+y[i]] == 1)chess[row+x[i]][col+y[i]] = 0;elsechess[row+x[i]][col+y[i]] = 1;}return; } void dfs(int row,int col, int deep)//深搜固定步數看能否同色 {if(deep == step){flag = judge(chess);return;}if(flag || row == 5)return;flip(row,col);if(col < 4)dfs(row,col+1,deep+1);elsedfs(row+1,1,deep+1);flip(row,col);//不符合就翻回之前的狀態if(col < 4)dfs(row,col+1,deep);elsedfs(row+1,1,deep);return; } int main() {char temp;int i, j;memset(chess,0,sizeof(chess));for(i = 1; i <= 4; i++){for(j = 1; j <= 4; j++){cin >>temp;if(temp == 'b')chess[i][j] = 1;}}for(step = 0; step <= 16; step++){//對每一步進行枚舉（Enum）dfs(1,1,0);if(flag)break;}if(flag)cout<<step<<endl;elsecout<<"Impossible"<<endl;return 0; }

轉載于:https://www.cnblogs.com/jzdwajue/p/6815646.html

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