UVA11205The broken pedometer
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UVA-11205
题意:给出n个p位的2进制串,求最少保留多少位可以区分这些2进制串。
解题思路:枚举子集。1 -> 2^p 中的每个数字用二进制表示都是一种子集的获取方式。把2进制串转成10进制,用 & 运算的取保留下的值,判断是否都不一样,都不一样计算位数,取最小的一个。
/************************************************************************* > File Name: UVA-11205.cpp > Author: Narsh > > Created Time: 2016年07月22日 星期五 15时36分01秒 ************************************************************************/#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;int s[300];bool pd[400000];int n,m,k,l,t,tot;bool flag;int main() { //freopen("xx.in","r",stdin); scanf("%d",&t); memset(pd,true,sizeof(pd)); while (t--) { scanf("%d%d",&m,&n); for (int i = 1; i <= n; i++){ l=0; for (int j = 1; j <= m; j++){ scanf("%d",&k); l=l*2+k; } s[i]=l; } tot=16; for (int r = 0; r < (1<<m); r++) { flag = true; int temp; for (int i = 1; i <= n; i++) { temp = s[i] & r; if (pd[temp]) pd[temp]=false; else { flag = false; break; } } if (flag) { l = 0; for (int i = 0; i < m; i++) if ( r & 1 << i) l++; if (l < tot) tot = l; } for (int i = 1; i <= n; i++){ temp = s[i] & r; pd[temp]=true; } } printf("%d\n",tot); }}
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