ZOJ 3675 Trim the Nails(bfs+状态压缩搜索)

Trim the Nails

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Robert is clipping his fingernails. But the nail clipper is old and the edge of the nail clipper is potholed.

The nail clipper's edge is N millimeters wide. And we use N characters('.' or '*') to represent the potholed nail clipper. '.' represents 1 bad millimeter edge, and '*' represents 1 good millimeter edge.(eg. "*****" is a 5 millimeters nail clipper with the whole edge good. "***..." is a 6 millimeters nail clipper with half of its edge good and half of its edge bad.)

Notice Robert can turn over the clipper. Turning over a "**...*"-nail clipper will make a "*...**"-nail clipper.

One-millimeter good edge will cut down Robert's one-millimeter fingernail. But bad one will not. It will keep the one-millimeter unclipped.

Robert's fingernail is M millimeters wide. How many times at least should Robert cut his fingernail?

Input

There will be multiple test cases(about 15). Please process to the end of input.

First line contains one integer N.(1≤N≤10)

Second line contains N characters only consists of '.' and '*'.

Third line contains one integer M.(1≤M≤20)

Output

One line for each case containing only one integer which is the least number of cuts. If Robert cannot clipper his fingernail then output -1.

Sample Input

8****..**46*..***7

Sample Output

12

Hint

We use '-' to present the fingernail.For sample 1:fingernail:----nail clipper:****..**Requires one cut.For sample 2:fingernail:-------nail clipper:*..***nail clipper turned over: ***..*Requires two cuts.题目大意：罗伯特的指甲长为m，他有一把刀，刀的长度为n（有'*'和'.'构成），'*'表示刀在该位置处是好的，能削减m的一个单位长度,'.'表示刀在该处是坏的，不能削减任何指甲。给定n和刀,还有m，求最少他需要多少次修剪才能将指甲削减好（长度为m的指甲全部被削减完）。如果，不能削减完，则输出-1.思路： 状态压缩搜索+bfs.前者是用来枚举状态空间，后者是用来求最短的修剪次数。状态压缩：用二进制位表示每一位被修剪的状态，1表示该位已经被修剪，0表示没有被修建。那么最终的状态就是m个1所表示的二进制数，即(1<<m)-1;
/*    @author : liuwen*/#include <iostream>#include <cstdio>#include <cstring>#include <climits> //INT_MAX INT_MIN LONG_LONG_MAX LONG_LONG_MIN#include <cstdlib>#include <queue>#include <stack>#include <map>#include <vector>#include <cmath>#include <algorithm>using namespace std;const int maxn=(1<<21)-1;struct Node{    int status;    int step;    Node(int status=0,int step=0):status(status),step(step){};}st,ed;int vis[maxn],minStep,n,m;int up,down;int toInt(string& s){    int sum=0;    string::size_type i=0;    for(i=0;i<s.size();i++){        if(s[i]!='0')   break;    }    for(;i<s.size();i++){        sum=(sum<<1)|(s[i]=='1');    }    return sum;}int bfs(Node st,Node ed){    int i,j,state,nowStatus;    memset(vis,0,sizeof(vis));    queue<Node>que;    que.push(st);    vis[st.status]=1;    while(!que.empty()){        Node tmp=que.front();        que.pop();        if(tmp.status==ed.status){            minStep=tmp.step;            return minStep;        }        for(j=0,state=up;j<2;j++,state=down){            for(i=0;i<=n;i++){                int tmpStatus=(state>>i); //枚举向右移i位,即                tmpStatus=(tmpStatus|tmp.status);//在父节点的指甲状态上的修建后                nowStatus=tmpStatus&((1<<m)-1); //子节点修建后的指甲状态                if(!vis[nowStatus]){                    vis[nowStatus]=1;                    que.push(Node(nowStatus,tmp.step+1));                }            }        }    }    return -1;}int main(){    //freopen("in.txt","r",stdin);    string str;    while(cin>>n>>str>>m){        for(string::size_type i=0;i<str.size();i++){            if(str[i]=='*') str[i]='1';            else    str[i]='0';        }        up=toInt(str);        reverse(str.begin(),str.end());        down=toInt(str);        up=up<<m;        down=down<<m;        n=n+m;        st=Node(0,0);        ed=Node((1<<m)-1,0);        cout<<bfs(st,ed)<<endl;    }    return 0;}