POJ 3740 Easy Finding

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

dfs+剪枝

#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<vector>
#include<cmath>
#include<string>
using namespace std;
const int maxn=305;
int n,m,a[20][maxn],b[20][20],c[20],d[20],cnt;
bool flag;

void dfs(int x)
{
	if (cnt==m) {flag=true; return;}
	if (flag||x>=n) return ;
	dfs(x+1);
	if (c[x]==0)
	{
		c[x]=1;
		int e[20]={0};
		for (int i=0;i<n;i++) 
			if (b[x][i]&&c[i]==0) {c[i]=1; e[i]=1;}
		cnt+=d[x];
		dfs(x+1);
		for (int i=0;i<n;i++) if (e[i]) c[i]=0;
		c[x]=0;
		cnt-=d[x];
	}
}

int main()
{
	while (cin>>n>>m)
	{
		memset(b,0,sizeof(b));
		memset(d,0,sizeof(d));
		memset(c,0,sizeof(c));
		for (int i=0;i<n;i++)
			for (int j=0;j<m;j++) scanf("%d",&a[i][j]);
		for (int i=0;i<n;i++)
			for (int j=0;j<m;j++)
			{
				if (a[i][j]) d[i]++;
				for (int k=i+1;k<n;k++)
					if (a[i][j]&&a[k][j]) b[i][k]=1;
			}
		flag=false;
		cnt=0;
		dfs(0);
		if (flag) printf("Yes, I found it\n");
		else printf("It is impossible\n");
	}
	return 0;
}
dlx
<pre name="code" class="cpp">#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const ll maxn=10005;int n,m;struct node{	int l,r,u,d;	int x,y;	node(int l=0,int r=0,int u=0,int d=0,int x=0,int y=0):	l(l),r(r),u(u),d(d),x(x),y(y){}};struct DLX{	node e[maxn];	int n,sz,num;	int ans[maxn],cnt[maxn];	void begin(int x)	{		sz=(n=x)+1;		for (int i=0;i<=n;i++)		{			e[i]=node((i+n)%(n+1),(i+1)%(n+1),i,i,0,i);		}		memset(cnt,0,sizeof(cnt));	}	void insert(int row)	{		int now=sz,x;		for (int i=1;i<=n;i++)		{			scanf("%d",&x);			if (x) 			{				e[sz]=node(sz-1,sz+1,e[i].u,i,row,i);				e[i].u=e[e[i].u].d=sz;				cnt[i]++;	sz++;			}		}		if (sz>now) e[sz-1].r=now,e[now].l=sz-1;	}	void remove(int x)	{		e[e[x].l].r=e[x].r;		e[e[x].r].l=e[x].l;		for (int i=e[x].d;i!=x;i=e[i].d)			for (int j=e[i].r;j!=i;j=e[j].r)			{				e[e[j].d].u=e[j].u;				e[e[j].u].d=e[j].d;				cnt[e[j].y]--;			}	}	void resume(int x)	{			e[e[x].l].r=x;		e[e[x].r].l=x;		for (int i=e[x].d;i!=x;i=e[i].d)			for (int j=e[i].r;j!=i;j=e[j].r)			{				e[e[j].d].u=j;				e[e[j].u].d=j;				cnt[e[j].y]++;			}	}	bool dfs(int x)	{		if (!e[0].r){ num=x; return true;}		int c=e[0].r;		for (int i=e[0].r;i!=0;i=e[i].r)			if (cnt[i]<cnt[c]) c=i;		remove(c);		for (int i=e[c].d;i!=c;i=e[i].d)		{			ans[x]=e[i].x;			for (int j=e[i].r;j!=i;j=e[j].r)			remove(e[j].y);			if (dfs(x+1)) return true;			for (int j=e[i].r;j!=i;j=e[j].r)			resume(e[j].y);		}		resume(c);		return false;	}	bool solve()	{		return dfs(0);	}}dlx;int main(){	while (~scanf("%d%d",&n,&m))	{		dlx.begin(m);		for (int i=1;i<=n;i++) dlx.insert(i);		if (dlx.solve()) printf("Yes, I found it\n");		else printf("It is impossible\n");	}	return 0;}