POJ3740 Easy Finding(Dancing Links)

G - Easy Finding

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status

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  Nintegers 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

题意：求任取n行能否使每列有且仅有一个1，

分析：算是数据结构加搜索，由著名的Dancing links解，粘一个 Dancing links详解地址：dancing links详解

我的ac代码(也是看了好长时间上面的文章自己又整理做的，先存了作为模板)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=350;
const int MAXM=350;
const int MAXX=MAXN*MAXM+MAXM;
int L[MAXX],R[MAXX],U[MAXX],D[MAXX],H[MAXX],C[MAXX];//UDLR分别为上下左右，H为该节点的行数，C为列数
int hh[MAXN],cc[MAXM];//hh为每行当前节点，cc为每列当前节点，建图使用
int ans[MAXN],n,m,cnt,head;

void init()
{
	int i;head=0;
	for(i=1;i<=n;i++)
		hh[i]=-1;
	for(i=0;i<=m;i++)
	{
		cc[i]=i;
		L[i+1]=i;
		R[i]=i+1;
		U[i]=D[i]=i;
		H[i]=0;
		C[i]=i;
	}
	L[0]=m;
	R[m]=0;
	cnt=m+1;
}
void insert(int a,int b)
{
	H[cnt]=a,C[cnt]=b;
	if(hh[a]==-1) L[cnt]=R[cnt]=cnt;
	else
	{
		L[cnt]=hh[a];
		L[R[hh[a]]]=cnt;
		R[cnt]=R[hh[a]];
		R[hh[a]]=cnt;
	}
	U[cnt]=cc[b];
	U[D[cc[b]]]=cnt;
	D[cnt]=D[cc[b]];
	D[cc[b]]=cnt;
	hh[a]=cnt;
	cc[b]=cnt;
	cnt++;
}
void remove(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])
		{
			D[U[j]]=D[j];
			U[D[j]]=U[j];
		}
}
void resume(int c)
{
	L[R[c]]=c;
	R[L[c]]=c;
	for(int i=U[c];i!=c;i=U[i])
		for(int j=R[i];j!=i;j=R[j])
		{
			U[D[j]]=j;
			D[U[j]]=j;
		}
}
bool dance(int k)
{
	int i,j,c=R[head];
	if(c==head)/*Showpath*/return true;
	remove(c);
	for(i=D[c];i!=c;i=D[i])
	{
		ans[k]=H[i];
		for(j=R[i];j!=i;j=R[j])remove(C[j]);
		if(dance(k+1))return true;
		for(j=L[i];j!=i;j=L[j])resume(C[j]);
	}
	resume(c);
	return false;
}
int main()
{
	int i,j,x;
	//freopen("123.txt","r",stdin);
	while(~scanf("%d%d",&n,&m))
	{
		init();
		for(i=1;i<=n;i++)
			for(j=1;j<=m;j++)
			{
				scanf("%d",&x);
				if(x)insert(i,j);
			}
		if(dance(0))printf("Yes, I found it\n");
		else printf("It is impossible\n");
	}
	return 0;
}