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DancingLinks[HUST_1017]

2013-03-06 12:11 197 查看
因为输出少了个\n WA了半天。 而且以下两种写法效率不一样。 有待研究

DancingLinks用于优化搜索，核心在于resume操作，可以快速恢复被删除的结点

1. 1468ms

/*http://acm.hust.edu.cn/problem.php?id=1017*/
/*DLX*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int MAXN = 200000;
const int MAXM = 1111;
struct DLX{
int L[MAXN],R[MAXN],U[MAXN],D[MAXN],C[MAXN],Col[MAXN],Row[MAXN];
int S[MAXM],Ans[MAXM];
int tot,len;
void build(int N,int M){
memset(S,0,MAXM*sizeof(int));
//construct the head and col head
for(int i=0;i<=M;i++){
U[i] = D[i] = i;
L[i] = i-1;
R[i] = i+1;
C[i] = i;
}
L[0] = M, R[M] = 0; tot = M;
for(int i=1;i<=N;i++){
int K,T; scanf("%d",&K);
for(int j=1;j<=K;j++){
scanf("%d",&T);
int node = ++tot;
C[node] = Col[node] = T;
Row[node] = i;
S[T]++;//维护列链表节点数
//维护UD两个指针
D[node] = D[C[T]];
D[C[T]] = node;
U[D[node]] = node;
U[D[C[T]]] = C[T];
C[T] = node; //更新到尾
//维护LR两个指针
if(j==1) L[node] = node+K-1;
else L[node] = node-1;
if(j==K) R[node] = node-K+1;
else R[node] = node+1;
}
}
for(int i=1;i<=M;i++){
C[i] = i;
}//还原列头指针
}
void debug(){
for(int i=R[0];i!=0;i=R[i]){
printf("C%d: ",i); int r = 1;
for(int j=D[i];j!=i;j=D[j]){
int ri = Row[j]; //cout<<"ri "<<ri<<" "<<r<<endl;
while(ri>r){
r++; printf("0 ");
}
printf("1 "); r++;
//printf("j: %d\n",j);
//system("pause");
}
printf("\n");
}
}
void remove(const 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[C[j]]--;
}
}
}
void resume(const int &c){
L[R[c]] = c;
R[L[c]] = c;
for(int i=D[c];i!=c;i=D[i]){
for(int j=R[i];j!=i;j=R[j]){
U[D[j]] = j;
D[U[j]] = j;
S[C[j]]++;
}
}
}
bool dfs(const int &k){
if(R[0]==0){
len = k;
return true;
}
int s(0x7fffffff),c;
for(int i=R[0];i!=0;i=R[i]){
if(S[i]<s){
c = i;
s = S[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(C[j]);
}
if(dfs(k+1)){
return true;
}
for(int j=R[i];j!=i;j=R[j]){
resume(C[j]);
}
}
resume(c);
return false;
}
void solve(){
if(dfs(0)){
printf("%d",len);
for(int i=0;i<len;i++){
printf(" %d",Ans[i]);
}
puts("");
}else{
puts("NO");
}
}
}dlx;
int main(){
int N,M;
while(~scanf("%d%d",&N,&M)){
dlx.build(N,M);
dlx.solve();
}
return 0;
}


2.328ms

/*http://acm.hust.edu.cn/problem.php?id=1017*/
/*DLX*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int MAXN = 2000000;
const int MAXM = 1111;
struct DLX{
int L[MAXN],R[MAXN],U[MAXN],D[MAXN],C[MAXN],Col[MAXN],Row[MAXN];
int S[MAXM],Ans[MAXM];
int tot,len;
void build(int N,int M){
memset(S,0,MAXM*sizeof(int));
//construct the head and col head
for(int i=0;i<=M;i++){
U[i] = D[i] = i;
L[i] = i-1;
R[i] = i+1;
C[i] = i;
}
L[0] = M, R[M] = 0; tot = M;
for(int i=1;i<=N;i++){
int K,T; scanf("%d",&K);
for(int j=1;j<=K;j++){
scanf("%d",&T);
int node = ++tot;
C[node] = Col[node] = T;
Row[node] = i;
S[T]++;//维护列链表节点数
//维护UD两个指针
D[node] = D[C[T]];
D[C[T]] = node;
U[D[node]] = node;
U[D[C[T]]] = C[T];
C[T] = node; //更新到尾
//维护LR两个指针
if(j==1) L[node] = node+K-1;
else L[node] = node-1;
if(j==K) R[node] = node-K+1;
else R[node] = node+1;
}
}
for(int i=1;i<=M;i++){
C[i] = i;
}//还原列头指针
}
void debug(){
for(int i=R[0];i!=0;i=R[i]){
printf("C%d: ",i); int r = 1;
for(int j=D[i];j!=i;j=D[j]){
int ri = Row[j]; //cout<<"ri "<<ri<<" "<<r<<endl;
while(ri>r){
r++; printf("0 ");
}
printf("1 "); r++;
//printf("j: %d\n",j);
//system("pause");
}
printf("\n");
}
}
void remove(const 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[C[j]]--;
}
}
}
void resume(const int &c){
for(int i=U[c];i!=c;i=U[i]){
for(int j=L[i];j!=i;j=L[j]){
U[D[j]] = j;
D[U[j]] = j;
S[C[j]]++;
}
}
L[R[c]] = c;
R[L[c]] = c;
}
bool dfs(const int &k){
if(R[0]==0){
len = k;
return true;
}
int s(0x7fffffff),c;
for(int i=R[0];i!=0;i=R[i]){
if(S[i]<s){
c = i;
s = S[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(C[j]);
}
if(dfs(k+1)){
return true;
}
for(int j=L[i];j!=i;j=L[j]){
resume(C[j]);
}
}
resume(c);
return false;
}
void solve(){
if(dfs(0)){
printf("%d",len);
for(int i=0;i<len;i++){
printf(" %d",Ans[i]);
}
puts("");
}else{
puts("NO");
}
}
}dlx;
int main(){
int N,M;
while(scanf("%d%d",&N,&M)!=EOF){
dlx.build(N,M);
dlx.solve();
}
return 0;
}
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