poj 3740 Easy Finding（Dancing Links 精確覆蓋）

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

3 3
0 1 0
0 0 1
1 0 0
4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0

Sample Output

Yes, I found it
It is impossible

題意：給出一個矩陣，問是否存在一些行使得每一列有且只有一個1。

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

AC代碼：

#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;
}