Flip Game dfs

2019-11-11 07:22:11

字體：大中小

來源：轉載

供稿：網友

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. InputThe input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.OutputWrite 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

bwwbbbwbbwwbbwwwSample Output4
解題報告
不要想太復雜就好，要明確如果有答案，那么答案小于等于8
#include<stdio.h>#include<string.h>bool map[6][6];int ox[]={0,0,0,1,-1};int oy[]={0,1,-1,0,0};int best;bool check(){    //end check    for(int j=1;j<=4;j++)        for(int k=1;k<=4;k++)            if(map[1][1]!=map[j][k])                return false;    return true;}void op(int X,int Y){    for(int i=0;i<5;i++){        int x=X+ox[i];        int y=Y+oy[i];        map[x][y]=!map[x][y];    }}void dfs(int t,int cnt){    if(check()){        if(cnt<best) best=cnt;        return ;    }    if(cnt>best||t>16) return ;    int X=t%4+1;    int Y=t/4+1;    //剪枝 加上該條件 78ms立馬變0ms    if(Y>1&&X>1&&map[X-1][Y-1]!=map[1][1]) return ;    dfs(t+1,cnt);    op(X,Y);    dfs(t+1,cnt+1);    op(X,Y);}int main(){    char str[5];    for(int i=1;i<=4;i++){        scanf("%s",str);        for(int j=1;j<=4;j++)            map[i][j]=str[j-1]=='b';    }    best=9;    dfs(0,0);    if(best<9) printf("%d/n",best);    else puts("Impossible");    return 0;}

上一篇：Apache Commons Compress

下一篇：通道