poj 2965 The Pilots Brothers' refrigerator[ 枚举 ]

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传送门：http://poj.org/problem?id=2965

思路：二进制枚举，递归输出路径。G++ 900+Ms勉强过，C++超时。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#define INF 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long LL;const int N=100007;int vis[N];int step[N][4];int change(int state,int i,int j){    for(int k=0;k<4;k++)    {        state=state^1<<(4*k+j);        state=state^1<<(4*i+k);    }    state=state^1<<(i*4+j);    return state;}void output(int state){    if(step[state][0]==0)        return;    else{        output(step[state][1]);        printf("%d %d\n",step[state][2],step[state][3]);    }}void bfs(int state){    queue<int> que;    que.push(state);    while(!que.empty())    {        int cur=que.front();        que.pop();        if(cur==65535)        {            printf("%d\n",step[65535][0]);            output(65535);            break;        }        for(int i=0;i<4;i++){            for(int j=0;j<4;j++){                int tmp=change(cur,i,j);                if(!vis[tmp])                {                    vis[tmp]=1;                    step[tmp][0]=step[cur][0]+1;                    step[tmp][1]=cur;                    step[tmp][2]=i+1;                    step[tmp][3]=j+1;                    que.push(tmp);                }            }        }    }}int main(){    //mem(vis,0);    //mem(step,0);    int state=0;    for(int i=0;i<16;i++)    {        char c;        cin>>c;        if(c=='-')        {            state=state|1<<i;        }    }    vis[state]=1;    bfs(state);    return 0;}

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