POJ基本算法bfs1753,2965

POJ1753

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: 

1. Choose any one of the 16 pieces. 
   
2. 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

16位数存储，每次操作即为异或。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;bool vis[65540];int p[65540],v[65540],ope[16],l,r;int bfs(int ans){    if(ans==0||ans==(1<<16)-1)        return 0;    memset(vis,0,sizeof(vis));    l=r=0;    vis[ans]=1,p[r++]=ans,v[0]=0;    while(l<r){        for(int i=0;i<16;i++){            int num=p[l]^ope[i];            if(vis[num])                continue;            if(num==0||num==(1<<16)-1)                return v[l]+1;            v[r]=v[l]+1;            p[r++]=num;            vis[num]=1;        }        l++;    }    return -1;}int main(){char s[4][5];    for(int i=0;i<4;i++)        for(int j=0;j<4;j++){            int num=4*i+j;            ope[num]=1<<num;            if(i-1>=0)                ope[num]+=1<<(num-4);            if(j-1>=0)                ope[num]+=1<<(num-1);            if(j+1<4)                ope[num]+=1<<(num+1);            if(i+1<4)                ope[num]+=1<<(num+4);        }    scanf("%s%s%s%s",s[0],s[1],s[2],s[3]);        int ans=0;        for(int i=3;i>=0;i--)            for(int j=3;j>=0;j--){                ans<<=1;                if(s[i][j]=='b')                    ans++;            }        int a=bfs(ans);        if(a==-1)            printf("Impossible\n");        else            printf("%d\n",a);return 0;}

POJ2965

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+-----------+--

Sample Output

61 11 31 44 14 34 4

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;bool vis[65540];struct op{    int a[16];}h[65540];//操作过程int p[65540],v[65540],ope[16],l,r;//节点 步数 操作int bfs(int ans){    if(!ans) return 0;    l=r=0;    for(int i=0;i<16;i++)        h[0].a[i]=0;    vis[ans]=1;    v[r]=0;    p[r++]=ans;    while(l<r){        for(int i=0;i<16;i++){            int num=ope[i]^p[l];            if(!vis[num]){                vis[num]=1;                v[r]=v[l]+1;                h[r]=h[l];                h[r].a[v[r]]=i;                p[r++]=num;                if(!num)                    return r-1;            }        }        l++;    }    return -1;}int main(){char s[4][5];int x[4],y[4];x[0]=(1<<4)-1;y[0]=1+(1<<4)+(1<<8)+(1<<12);for(int i=1;i<4;i++)        x[i]=x[i-1]<<4,y[i]=y[i-1]<<1;    for(int i=0;i<4;i++)        for(int j=0;j<4;j++)            ope[4*i+j]=x[i]|y[j];    while(scanf("%s%s%s%s",s[0],s[1],s[2],s[3])!=EOF){        int ans=0;        for(int i=3;i>=0;i--)            for(int j=3;j>=0;j--){                ans<<=1;                if(s[i][j]=='+')                    ans++;            }        int n=bfs(ans);        printf("%d\n",v[n]);        for(int i=1;i<=v[n];i++)        {            int mid=h[n].a[i];            printf("%d %d\n",mid/4+1,mid%4+1);        }    }return 0;}

bfs，第i个状态得来的操作步骤存储在h[i]的op数组里。（最多不会超过16个操作步骤）