PKU 2724 - Purifying Machine (最小边覆盖)

题意

有一些奶酪被污染了,现在要清洗一下。

有*的代表0 1都可以。

问最少要清洗几次。

思路

学习了判断两个二进制是不是只差一位。

对只差一位的序列建边,因为正反都建了一遍,所以最大匹配多了一倍。

最小边覆盖 = V - 最大匹配

代码

#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 2000 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
struct EDGE
{
    int from, to, cap;
};
 
int match[MAXN], ele[MAXN];
vector<int> G[MAXN];
 
struct HUNGARY
{
    int vis[MAXN];
    int solve(int nrem)
    {
        int res = 0;
        MS(match, 0);
        for (int i = 0; i < nrem; i++)
        {
            MS(vis, 0);
            res += DFS(i);
        }
        return res;
    }
 
    int DFS(int u)
    {
        for (int i = 0; i < SZ(G[u]); i++)
        {
            int v = G[u][i];
            if (!vis[v])
            {
                vis[v] = 1;
                if (!match[v] || DFS(match[v]))
                {
                    match[v] = u;
                    return 1;
                }
            }
        }
        return 0;
    }
}hun;
 
int main()
{
    //ROP;
    int n, nope, i, j;
    while (~scanf("%d%d", &n, &nope), n + nope)
    {
        int cur = 0;
        char cmd[15];
        while (nope--)
        {
            scanf("%s", cmd);
            int sum = 0, pos = -1;
            for (i = 0; i < n; i++)
            {
                if (cmd[i] == '*') pos = i;
                else sum |= ((cmd[i] - '0') << i);
            }
            ele[cur++] = sum;
            if (pos != -1) ele[cur++] = (sum | (1 << pos));
        }
        sort(ele, ele + cur);
        int len = unique(ele, ele + cur) - ele;
        for (i = 0; i <= len; i++) G[i].clear();
        for (i = 0; i < len; i++)
            for (j = i + 1; j < len; j++)
            {
                int tmp = ele[i] ^ ele[j];
                if (tmp && ((tmp & (tmp - 1)) == 0))
                {
                    G[i].PB(j);
                    G[j].PB(i);
                }
            }
        printf("%d\n", len - hun.solve(len) / 2);
    }
    return 0;
}

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