跟上题差不多,还是一个01矩阵,问是否可以选出若干行使得每列恰有一个1。模型都给建好了,直接建表搜就可以了..
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn=320*320;
struct DLX
{
int col[maxn],row[maxn],ans[maxn];
int S[maxn],H[maxn];
int size,cnt;
int U[maxn],R[maxn],D[maxn],L[maxn];
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])
{
U[D[j]]=U[j];
D[U[j]]=D[j];
--S[col[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;
++S[col[j]];
}
}
}
bool dance(int k)
{
int c=R[0];
if (c==0)
{
cnt=k;
return true;
}
for (int i=R[0]; i!=0; i=R[i])
{
if (S[c]>S[i]) c=i;
}
remove(c);
for (int i=D[c]; i!=c; i=D[i])
{
ans[k]=row[i];
for (int j=R[i]; j!=i; j=R[j]) remove(col[j]);
if (dance(k+1)) return true;
for (int j=R[i]; j!=i; j=R[j]) resume(col[j]);
}
resume(c);
return false;
}
void init(int n)
{
for (int i=0; i<=n; i++)
{
S[i]=0;
L[i]=i-1;
R[i]=i+1;
U[i]=D[i]=i;
}
L[0]=n;
R[n]=0;
size=n+1;
memset(H,-1,sizeof H);
}
void link(int x,int y)
{
++S[col[size]=y];
row[size]=x;
D[size]=D[y];
U[D[y]]=size;
U[size]=y;
D[y]=size;
if (H[x]<0) H[x]=L[size]=R[size]=size;
else
{
R[size]=R[H[x]];
L[R[H[x]]]=size;
L[size]=H[x];
R[H[x]]=size;
}
size++;
}
}dlx;
int n,m,k,num;
int main()
{
// freopen("in.txt","r",stdin);
while (~scanf("%d%d",&m,&n))
{
dlx.init(n);
for (int i=1; i<=m; i++)
for (int j=1; j<=n; j++)
{
scanf("%d",&k);
if (k)
{
dlx.link(i,j);
}
}
if (dlx.dance(0))
{
puts("Yes, I found it");
}
else puts("It is impossible");
}
return 0;
}