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;
}