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; }