poj2724 Purifying Machine

每个奶酪上面n位二进制，含有0，1，＊三种符号，现在有m个奶酪，我们需要用一个机器来clear这些奶酪，clear的条件就是机器上的n位二进制与奶酪上面的二进制相同，但是＊可以代替0或1，所以这样的话只要两个奶酪的只有一个位置不同的话，还是可以用同一个改变clear两个奶酪了，所以这样就建边。然后就是最小边覆盖。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1 << 11;int g[N][N];int num[N*2];int n, m;int top;char s[20];void input() {top = 0;for (int i = 1;i <= m;++i) {scanf("%s",s);int ans = 0;for (int j = 0;j < n;++j) {if (s[j] == '*') ans = ans * 2 + 0;else ans = ans * 2 + s[j] - '0';}num[++top] = ans;ans = 0;for (int j = 0;j < n;++j) {if (s[j] == '*') ans = ans * 2 + 1;else ans = ans * 2 + s[j] - '0';}num[++top] = ans;}sort(num + 1,num + 1 + top);// for (int i = 1;i <= top;++i)// printf("%4d",num[i]);int up = 1;for (int i = 2;i <= top;++i) {if (num[i] != num[up]) {num[++up] = num[i];}}top = up;// cout << "top = " << top << endl;memset(g, 0,sizeof g);for (int i = 1;i <= top;++i) {for (int j = i + 1;j <= top;++j) {int t = num[i] ^ num[j];if (t && (t&(t-1))==0)g[i][j] = g[j][i] = 1;}}// for (int i = 1;i <= top;++i) {// for (int j = 1;j <= top;++j)// printf("%2d",g[i][j]);// puts("");// }}int cx[N], cy[N];bool vis[N];bool dfs(int u) {for (int i = 1;i <= top;++i) {if (!vis[i] && g[u][i]) {vis[i] = true;if (cy[i] == 0 || dfs(cy[i])) {cy[i] = u;cx[u] = i;return true;}}}return false;}int hungary() {memset(cx, 0,sizeof cx);memset(cy, 0,sizeof cy);int ans = 0;for (int i = 1;i <= top;++i) {memset(vis, false,sizeof vis);if (dfs(i)) ans++;}return ans;}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(~scanf("%d%d",&n,&m) && (n + m)) {input();int ans = hungary();// cout << top << endl;printf("%d\n",top - ans/2);}return 0;}