poj 3740 Easy Finding（Dancing Links 精确覆盖）

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#Dancing Links #精确覆盖

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本文介绍了一种使用Dancing Links算法解决特定矩阵问题的方法，该问题要求找到某些行组合，使每列恰好包含一个1。通过详细的AC代码示例，展示了如何实现这一算法。

Easy Finding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16128		Accepted: 4321

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

题意：给出一个矩阵，问是否存在一些行使得每一列有且只有一个1。

思路：Dancing Links。详见http://www.cnblogs.com/grenet/p/3145800.html

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <bitset>
#include <queue>
#define ll long long
using namespace std;

const int maxn = 6005;
const int INF = 1e9;

int n, m, cnt, head;
int L[maxn], R[maxn], U[maxn], D[maxn], S[maxn], C[maxn], H[maxn];
inline void add_link(int i, int j){
    C[++cnt] = j;
    S[j]++;
    D[cnt] = j;
    U[cnt] = U[j];
    if(H[i]) R[cnt] = H[i], L[cnt] = L[H[i]];
    else R[cnt] = L[cnt] = cnt;
    H[i] = cnt;
    U[D[cnt]] = cnt;
    D[U[cnt]] = cnt;
    R[L[cnt]] = cnt;
    L[R[cnt]] = cnt;
}
void remove(int c){
    R[L[c]] = R[c];
    L[R[c]] = L[c];
    for(int i = D[c]; i != c; i = D[i])
    for(int j = R[i]; j != i; j = R[j])
    {
        U[D[j]] = U[j];
        D[U[j]] = D[j];
        S[C[j]]--;
    }
}
void resume(int c){
    for(int i = U[c]; i != c; i = U[i])
    for(int j = L[i]; j != i; j = L[j])
    {
        D[U[j]] = j;
        U[D[j]] = j;
        S[C[j]]++;
    }
    L[R[c]] = R[L[c]] = c;
}
bool dance(){
    if(R[head] == head)
    {
        puts("Yes, I found it");
        return true;
    }
    int s = INF, c;
    for(int i = R[head]; i != head; i = R[i])
    if(S[i] < s) s = S[c = i];
    remove(c);
    for(int i = D[c]; i != c; i = D[i])
    {
        for(int j = R[i]; j != i; j = R[j]) remove(C[j]);
        if(dance()) return true;
        for(int j = L[i]; j != i; j = L[j]) resume(C[j]);
    }
    resume(c);
    return false;
}
int main()
{
    int c;
    head = 0;
    while(~scanf("%d%d", &n, &m))
    {
        cnt = m;
        for(int i =- 0; i <= m; i++)
        {
            C[i] = U[i] = D[i] = i;
            L[i + 1] = i;
            R[i] = i + 1;
            S[i] = 0;
        }
        L[0] = m, R[m] = 0;
        for(int i = 1; i <= n; i++)
        {
            H[i] = 0;
            for(int j = 1; j <= m; j++)
            {
                c = getchar();
                while(!isdigit(c)) c = getchar();
                if(c == '1') add_link(i, j);
            }
        }
        if (!dance()) puts("It is impossible");
    }
    return 0;
}