[UVA11205] The broken pedometer, train of thought + Code, may be The simplest, most understandable, and most efficient code.

# Include <cstdlib> # include <cstdio> # include <cstring> # include <algorithm> # include <queue> using namespace std;/** Problem: UVA11205-The broken pedometer Begin Time: 28th/Mar/2012 a.m. finish Time: 29th/Mar/2012 a.m. last Time: About 6 hours. state: 992440711205The broken pedometerAcceptedC ++ 2.6842012-03-28 18:11:32 knodge DGE: BFS, No pruning, and subset generation. Is there a small State compression? Exp ..: WA * 4: The four Wrong Answer gave a mistake to write BFS as DFS. It shouldn't be WA * 4: I am not capable of talking about the last four WA. Boundary, boundary! Tmp. depth + 1 <= col is written as tmp. depth + 1 <col, And a combination is missing! An idiot! When writing a program. Operation. Thought: replace each line with an integer like 10110101 (because it can only be 2 ^ 15 at most, it does not need bitwise Operations, otherwise it is a strange problem) to open an array isSel [], if the ith column is selected, isSel [I] = true; then the selected number of columns is determined based on isSel, And the subsets should be extracted from those columns. In fact, bit operations can be faster. Summerize: 1. think carefully about the search process, the status transfer method, whether there are status nodes not expanded 2. think about the boundary. The most important question is whether there is a problem. I can see it clearly before writing it. Generally, there are shortest, minimum, and minimum BFS words. if a tree is searched, no backtracking is required. If a graph is searched, backtracking is required. */Const int MAXN = 200; const int MAXP = 50; int puzzle [MAXN] [MAXP]; int subset [MAXN]; bool isSel [MAXN]; int minN; bool isFound; struct node {int depth; bool isSel [MAXN] ;}; node tmp; queue <node> que; int comp (const void * a, const void * B) {return (* (int *) a-* (int *) B);} bool check (int row, int col) {int cnt = 0; memset (subset, 0, sizeof (subset); for (int I = 0; I <col; I ++) {if (isSel [I] = true) cnt ++;} (Int j = 0; j <row; j ++) {for (int I = 0; I <col; I ++) {if (isSel [I] = true) {subset [j] = subset [j] * 10 + puzzle [j] [I] ;}} qsort (subset, row, sizeof (int), comp); int I = 0; int j = I + 1; while (j <row) {if (subset [I] = subset [j]) {return false;} I ++, j ++;} if (minN> cnt) {minN = cnt ;} return true;} int Solve (int row, int col, int now) {node tmp, tmp1, k; while (! Que. empty () {tmp = que. front (); que. pop (); memcpy (isSel, tmp. isSel, sizeof (isSel); tmp1 = tmp; check (row, col); if (tmp. depth + 1 <= col) {tmp1.depth = tmp. depth + 1; tmp1.isSel [tmp1.depth-1] = false; que. push (tmp1); tmp1.isSel [tmp1.depth-1] = true; que. push (tmp1) ;}}return minN ;}int main () {# ifndef ONLINE_JUDGE freopen ("B: \ acm \ ultraviolet \ UVA11205 \ input.txt ", "r", stdin); freopen ("B: \ acm \ ultraviolet \ uva11 205 \ output.txt "," w ", stdout); # endif int T, P, N, k; // while (scanf (" % d ", & T )! = EOF) scanf ("% d", & T); {for (int t = 1; t <= T; t ++) {scanf ("% d", & P, & N); memset (puzzle, 0, sizeof (puzzle); memset (subset, 0, sizeof (subset); memset (isSel, 0, sizeof (isSel); for (int I = 0; I <N; I ++) {for (int j = 0; j <P; j ++) {scanf ("% d", & puzzle [I] [j]) ;}} minN = 200; tmp. depth = 1; tmp. isSel [0] = true; que. push (tmp); tmp. depth = 1; tmp. isSel [0] = false; que. push (tmp); k = Solve (N, P, 0 ); Printf ("% d \ n", k); while (! Que. empty () {que. pop () ;}} return 0 ;}