Easy Finding POJ - 3740 （DLX）

显然这是一道dfs简单题

或许匹配也能做

然而用了dancing links

显然这也是一道模板题

好的吧

调了一上午 终于弄好了模板

Easy Finding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19052		Accepted: 5273

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

Source

POJ Monthly Contest - 2009.08.23, MasterLuo

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;

int S[maxn], head[maxn], vis[maxn];
int U[maxn], D[maxn], L[maxn], R[maxn];
int C[maxn], X[maxn];
int n, m, ans, ret;

void init()
{
    for(int i = ; i <= m; i++)
        D[i] = i, U[i] = i, R[i] = i + , L[i] = i - ;
    L[] = m, R[m] = ;
    mem(S, ), mem(head, -);
    ans = m + ;
}

void delc(int c)
{
    L[R[c]] = L[c], R[L[c]] = R[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 resc(int c)
{
    for(int i = U[c]; i != c; i = U[i])
        for(int j = L[i]; j != i; j = L[j])
            U[D[j]] = j, D[U[j]] = j, S[C[j]]++;
    L[R[c]] = c, R[L[c]] = c;
}

void add(int r, int c)
{
    ans++, S[c]++, C[ans] = c, X[ans] = r;
    D[ans] = D[c];
    U[ans] = c;
    U[D[c]] = ans;
    D[c] = ans;
    if(head[r] < ) head[r] = L[ans] = R[ans] = ans;
    else L[ans] = head[r], R[ans] = R[head[r]],L[R[head[r]]] = ans, R[head[r]] = ans;
}

bool dfs(int sh)
{
    if(!R[])
    {
        ret = sh;
        return true;
    }
    int c = R[];
    delc(c);
    for(int i = D[c]; i != c; i = D[i])
    {
        vis[sh] = i;
        for(int j = R[i]; j != i; j = R[j])
            delc(C[j]);
        if(dfs(sh + )) return true;
        for(int j = L[i]; j != i; j = L[j])
            resc(C[j]);
    }
    resc(c);
    return false;
}

int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        init();
        int tmp;
        for(int i = ; i <= n; i++)
            for(int j = ; j <= m; j++)
            {
                rd(tmp);
                if(tmp) add(i, j);
            }
        if(dfs())
            printf("Yes, I found it\n");
        else
            printf("It is impossible\n");

    }

    return ;
}