DancingLinks [HUST_1017]

Because the output is less than \ n WA for half a day. In addition, the efficiency of the two writing methods is different. To be studied, DancingLinks is used to optimize the search. The core is the resume operation, which can quickly restore the deleted Node 1. 1468 ms [cpp]/* 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] ++; // maintain the number of nodes in the column linked list // maintain the UD pointer D [node] = D [C [T]; D [C [T] = node; U [D [node] = node; U [D [C [T] = C [T]; C [T] = node; // update to the end // maintain two LR pointers 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;} // restore the column header pointer} 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", & N, & M) {dlx. build (N, M); dlx. solve ();} return 0;} 2.328 ms [cpp]/* 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] ++; // maintain the number of nodes in the column linked list // maintain the UD pointer D [node] = D [C [T]; D [C [T] = node; U [D [node] = node; U [D [C [T] = C [T]; C [T] = node; // update to the end // maintain two LR pointers 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;} // restore the column header pointer} 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; www.2cto.com while (scanf ("% d", & N, & M )! = EOF) {dlx. build (N, M); dlx. solve ();} return 0 ;}