2014 Super Training #6 G Trim the Nails --状态压缩+BFS

原题： ZOJ 3675 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3675

由m<=20可知，可用一个二进制数表示指甲的状态，最多2^20，初始状态为0，表示指甲都没剪，然后BFS找解，每次枚举剪刀的两个方向，枚举移动的位数进行扩展状态即可。

代码:

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

using namespace std;

#define N 10007



struct node

{

    int state,step;

    node(int _state,int _step)

    {

        state = _state;

        step = _step;

    }

    node(){}

};



int vis[1<<21];

int cut[2]; //两个方向

queue<node> que;

int n,m;



int bfs(int s)

{

    int i,j,k;

    memset(vis,0,sizeof(vis));

    while(!que.empty())

        que.pop();

    int E = (1<<m)-1;

    que.push(node(s,0));

    vis[s] = 1;

    while(!que.empty())

    {

        node tmp = que.front();

        que.pop();

        int state = tmp.state;

        int step = tmp.step;

        int tms = state;

        for(i=0;i<2;i++) //direction

        {

            for(j=0;j<n;j++)  //move

            {

                int end = ((cut[i]>>j) | tms) & E;  // &E : keep m bit

                if(vis[end])

                    continue;

                vis[end] = 1;

                if(end == E)

                    return step+1;

                que.push(node(end,step+1));

            }

            for(j=0;j<m;j++)

            {

                int to = ((cut[i]<<j) | tms) & E;

                if(vis[to])

                    continue;

                vis[to] = 1;

                if(to == E)

                    return step+1;

                que.push(node(to,step+1));

            }

        }

    }

    return -1;

}



int main()

{

    int i,j;

    char ss[13];

    while(scanf("%d",&n)!=EOF)

    {

        cut[0] = cut[1] = 0;

        scanf("%s",ss);

        for(i=0;i<=n;i++)

        {

            if(ss[i] == '*')

            {

                cut[0] |= (1<<i);

                cut[1] |= (1<<(n-1-i));

            }

        }

        scanf("%d",&m);

        if(cut[0] == 0)

        {

            puts("-1");

            continue;

        }

        printf("%d\n",bfs(0));

    }

    return 0;

}

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