POJ3740-Easy Finding

最新推荐文章于 2017-10-30 20:45:10 发布

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最新推荐文章于 2017-10-30 20:45:10 发布

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分类专栏： POJ ----dlx

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本文介绍了一个使用舞蹈链算法解决精确覆盖问题的经典案例——寻找特定矩阵中满足条件的行组合，确保每列恰好有一个1。通过详细解析源代码，展示了如何实现这一算法，并提供了具体的输入输出示例。

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Easy Finding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18660		Accepted: 5120

Description

Given a M× N matrix A. A _ij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.

Input

There are multiple cases ended by EOF. Test case up to 500.The first line of input is M, N ( M ≤ 16, N ≤ 300). The next M lines every line contains N integers separated by space.

Output

For each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line.

Sample Input

Sample Output

Yes, I found it
It is impossible

Source

POJ Monthly Contest - 2009.08.23, MasterLuo

题意：给你一个01矩阵，问能不能选出一些行，使得每一列都有且仅在一行的那一列为1

解题思路：舞蹈链的精确覆盖

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cctype>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int maxn = 300005;

int n, m, a[20][400];

struct DLX
{
	int L[maxn], R[maxn], U[maxn], D[maxn];
	int row[maxn], col[maxn], ans[maxn], sum[maxn];
	int n, m, num, cnt;
	void add(int k, int l, int r, int u, int d, int x, int y)
	{
		L[k] = l;   R[k] = r;   U[k] = u;
		D[k] = d;   row[k] = x;  col[k] = y;
	}
	void reset(int n, int m)
	{
		this->n = n;   this->m = m;
		for (int i = 0; i <= m; i++)
		{
			add(i, i - 1, i + 1, i, i, 0, i);
			sum[i] = 0;
		}
		L[0] = m, R[m] = 0, cnt = m + 1;
	}
	void insert(int x, int y)
	{
		int temp = cnt - 1;
		if (row[temp] != x)
		{
			add(cnt, cnt, cnt, U[y], y, x, y);
			U[D[cnt]] = cnt; D[U[cnt]] = cnt;
		}
		else
		{
			add(cnt, temp, R[temp], U[y], y, x, y);
			R[L[cnt]] = cnt; L[R[cnt]] = cnt;
			U[D[cnt]] = cnt; D[U[cnt]] = cnt;
		}
		sum[y]++, cnt++;
	}
	void remove(int k)
	{
		R[L[k]] = R[k], L[R[k]] = L[k];
		for (int i = D[k]; i != k; i = D[i])
			for (int j = R[i]; j != i; j = R[j])
			{
				D[U[j]] = D[j];
				U[D[j]] = U[j];
				sum[col[j]]--;
			}
	}
	void resume(int k)
	{
		R[L[k]] = k, L[R[k]] = k;
		for (int i = D[k]; i != k; i = D[i])
			for (int j = R[i]; j != i; j = R[j])
			{
				D[U[j]] = j;
				U[D[j]] = j;
				sum[col[j]]++;
			}
	}
	bool dfs(int k)
	{
		if (!R[0]) { num = k; return true; }
		int now = R[0];
		for (int i = now; i != 0; i = R[i])
			if (sum[now] > sum[i]) now = i;
		remove(now);
		for (int i = D[now]; i != now; i = D[i])
		{
			ans[k] = row[i];
			for (int j = R[i]; j != i; j = R[j]) remove(col[j]);
			if (dfs(k + 1)) return true;
			for (int j = L[i]; j != i; j = L[j]) resume(col[j]);
		}
		resume(now);
		return false;
	}
}dlx;

int main()
{
	while (~scanf("%d %d", &n, &m))
	{
		dlx.reset(n, m);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
			{
				scanf("%d", &a[i][j]);
				if (a[i][j]) dlx.insert(i, j);
			}
		if (dlx.dfs(0)) printf("Yes, I found it\n");
		else printf("It is impossible\n");
	}
	return 0;
}