UVa 1663 Purifying Machine (二分匹配)

题意：每一个01串中最多含有一个‘*’，‘*’既可表示0也可表示1，给出一些等长的这样的01串，问最少能用多少个这样的串表示出这些串。

如：000、010、0*1表示000、010、001、011，最少只需用00*、01*这两个即可表示出来。

析：因为最多只有一个星，所以每个串最多能代表两个串，所以就是要两两匹配的尽量多，也就是二分匹配喽，要注意，给的串可能会有重复的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
vector<string> v;
int V;
vector<int> G[maxn];
int match[maxn];
bool used[maxn];

void addEdge(int u, int v){
  G[u].push_back(v);
  G[v].push_back(u);
}

bool dfs(int v){
  used[v] = true;
  for(int i = 0; i < G[v].size(); ++i){
    int u = G[v][i], w = match[u];
    if(w < 0 || !used[w] && dfs(w)){
      match[v] = u;
      match[u] = v;
      return true;
    }
  }
  return false;
}

int solve(){
  int res = 0;
  memset(match, -1, sizeof match);
  for(int v = 0; v < V; ++v)  if(match[v] < 0){
    memset(used, 0, sizeof used);
    if(dfs(v))  ++res;
  }
  return res;
}

bool judge(int i, int j){
  int cnt = 0, k = 0;
  while(k < n){
    if(v[i][k] != v[j][k]) ++cnt;
    ++k;
  }
  return cnt == 1;
}

int main(){
  while(cin >> n >> m && m+n){
    v.clear();
    for(int i = 0; i < m ; ++i){
      string s;
      cin >> s;
      if(s.find('*') != string::npos){
        int pos = s.find('*');
        s[pos] = '1';
        v.push_back(s);
        s[pos] = '0';
      }
      v.push_back(s);
    }
    sort(v.begin(), v.end());
    V = unique(v.begin(), v.end()) - v.begin();
    for(int i = 0; i < V; ++i)  G[i].clear();
    for(int i = 0; i < V; ++i)
      for(int j = i+1; j < V; ++j)
        if(judge(i, j))  addEdge(i, j);
    printf("%d\n", V - solve());
  }
  return 0;
}