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
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;
}
