Zoj 3675 pressure DP

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 4918

Yesterday's ranking, I thought I was thinking, and my teammates told me to press DP and gave up directly. After the game, I checked my teams' code and searched the internet, the Code of teammates is actually the shortest. Worship ~~~~~~~

The idea is teammate a.l.

DP [s] = min (DP [s], DP [s '] + 1)

Among them, s' can be achieved through one-time nail cutting or reverse nail cutting.

As for the inner layer loop, 0-M is because ---- From the last bit of the Clipper, the last bit is moved from 0 to m, which is enough for all the situations.

#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const int MAXN = 22;const int INF =0x3f3f3f3f3f3f3f3f;int dp[1<<MAXN];int main(){    //IN("zoj3675.txt");    int up,down;    int n,m;    char str[MAXN];    while(~scanf("%d",&n))    {        scanf("%s%d",str,&m);        int len=strlen(str);        up=down=0;        for(int i=0;i<len;i++)        {            if(str[i]=='*')            {                up+=(1<<i);                down+=(1<<(n-i-1));            }        }        CL(dp,INF);        dp[(1<<m)-1]=0;        for(int i=(1<<m)-1;i>=0;i--)            for(int j=0;j<m;j++)            {                dp[i&~(up<<j)]=min(dp[i&~(up<<j)],dp[i]+1);                dp[i&~(down<<j)]=min(dp[i&~(down<<j)],dp[i]+1);            }        printf("%d\n",dp[0]>=INF?-1:dp[0]);    }    return 0;}

Zoj 3675 pressure DP