1017 - Exact cover
Time Limit: 15s Memory Limit: 128MB
Special Judge Submissions: 5851 Solved: 3092
- DESCRIPTION
- There is an N*M matrix with only 0s and 1s, (1 <= N,M <= 1000). An exact cover is a selection of rows such that every column has a 1 in exactly one of the selected rows. Try to find out the selected rows.
- INPUT
- There are multiply test cases. First line: two integers N, M; The following N lines: Every line first comes an integer C(1 <= C <= 100), represents the number of 1s in this row, then comes C integers: the index of the columns whose value is 1 in this row.
- OUTPUT
- First output the number of rows in the selection, then output the index of the selected rows. If there are multiply selections, you should just output any of them. If there are no selection, just output "NO".
- SAMPLE INPUT
-
6 7 3 1 4 7 2 1 4 3 4 5 7 3 3 5 6 4 2 3 6 7 2 2 7
- SAMPLE OUTPUT
-
3 2 4 6
- HINT
- SOURCE
- dupeng
- dancing link裸題，就不細說了。
-
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<ctime> #include<cmath> #include<algorithm> #include<set> #include<map> #include<vector> #include<string> #include<queue> using namespace std; #ifdef WIN32 #define LL "%I64d" #else #define LL "%lld" #endif #define MAXN 1100 #define MAXV MAXN*2 #define MAXE MAXV*2 #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3fLL #define MAXD 100000 typedef long long qword; inline int nextInt() { char ch; int x=0; bool flag=false; do ch=(char)getchar(),flag=(ch==‘-‘)?true:flag; while(ch<‘0‘||ch>‘9‘); do x=x*10+ch-‘0‘; while (ch=(char)getchar(),ch<=‘9‘ && ch>=‘0‘); return x*(flag?-1:1); } int n,m; struct DLX_t { static const int maxd=1000000; static const int maxn=1001; static const int maxm=1001; int L[maxd],R[maxd],U[maxd],D[maxd]; int rw[maxd]; int head; int topd; int chd[maxm]; int col[maxd]; int tt[maxm]; int n,m; vector<int> res; void init(int nn,int mm) { n=nn;m=mm; topd=0; memset(L,0,sizeof(L)); memset(R,0,sizeof(R)); memset(D,0,sizeof(D)); memset(U,0,sizeof(U)); memset(tt,0,sizeof(tt)); res.clear(); head=++topd; L[head]=R[head]=head; for (int i=1;i<=m;i++) { chd[i]=++topd; col[chd[i]]=i; rw[chd[i]]=0; R[chd[i]]=head; L[chd[i]]=L[head]; R[chd[i]]=head; L[R[chd[i]]]=chd[i]; R[L[chd[i]]]=chd[i]; U[chd[i]]=D[chd[i]]=chd[i]; } } void Add_row(int r,const vector<int> &vec) { int i; int nowh; int now; for (i=0;i<(int)vec.size();i++) { now=++topd; rw[now]=r; col[now]=vec[i]; tt[vec[i]]++; U[now]=U[chd[vec[i]]]; D[now]=chd[vec[i]]; D[U[now]]=now; U[D[now]]=now; } L[U[chd[vec[0]]]]=R[U[chd[vec[0]]]]=U[chd[vec[0]]]; nowh=U[chd[vec[0]]]; for (i=1;i<(int)vec.size();i++) { R[U[chd[vec[i]]]]=nowh; L[U[chd[vec[i]]]]=L[nowh]; L[R[U[chd[vec[i]]]]]=U[chd[vec[i]]]; R[L[U[chd[vec[i]]]]]=U[chd[vec[i]]]; } } void Finish() { vector<int> res2=res; sort(res2.begin(),res2.end()); printf("%d",(int)res2.size()); for (int i=0;i<(int)res2.size();i++) printf(" %d",res2[i]); printf("\n"); } void cover(int c) { int i,j; R[L[chd[c]]]=R[chd[c]]; L[R[chd[c]]]=L[chd[c]]; for (i=D[chd[c]];i!=chd[c];i=D[i]) { for (j=R[i];j!=i;j=R[j]) { tt[col[j]]--; U[D[j]]=U[j]; D[U[j]]=D[j]; } } } void resume(int c) { int i,j; R[L[chd[c]]]=chd[c]; L[R[chd[c]]]=chd[c]; for (i=D[chd[c]];i!=chd[c];i=D[i]) { for (j=R[i];j!=i;j=R[j]) { tt[col[j]]++; U[D[j]]=j; D[U[j]]=j; } } } bool dfs() { int now=head; if (L[now]==now) { Finish(); return true; } int bst=INF,bi=-1; int i,j; for (i=R[head];i!=head;i=R[i]) { if (tt[col[i]]<bst) { bst=tt[i]; bi=i; } } cover(col[bi]); for (i=D[bi];i!=bi;i=D[i]) { res.push_back(rw[i]); for (j=R[i];j!=i;j=R[j]) cover(col[j]); if (dfs())return true; res.pop_back(); for (j=R[i];j!=i;j=R[j]) resume(col[j]); } resume(col[bi]); return false; } }DLX; vector<int> vec; int main() { freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); int i,j,k; int x,y,z; while (~scanf("%d%d",&n,&m)) { DLX.init(n,m); for (i=1;i<=n;i++) { scanf("%d",&y); vec.clear(); for (j=1;j<=y;j++) { scanf("%d",&x); vec.push_back(x); } sort(vec.begin(),vec.end()); DLX.Add_row(i,vec); } if (!DLX.dfs()) printf("NO\n"); } return 0; }
时间： 2024-10-31 22:15:50