HDU 1581 square (DFS (deep first search ))

Square

Time Limit: 10000/5000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 6032 accepted submission (s): 1937

Problem descriptiongiven a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

Inputthe first line of input contains N, the number of test cases. each test case begins with an integer 4 <= m <= 20, the number of sticks. M integers follow; each gives the length of a stick-an integer between 1 and 10,000.

Outputfor each case, output a line containing "yes" if is possible to form a square; otherwise output "no ".

Sample Input

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

Sample output

yesnoyes

Question: Can a square be formed? Train of Thought: DFS algorithm method 1

Import Java. io. *; import Java. util. *; public class main {public static Boolean mark []; public static integer a []; public static Boolean boo; public static int edge, sum; public static int n, m; public static void main (string [] ARGs) {consumer SC = new consumer (New bufferedinputstream (system. in); printwriter PW = new printwriter (New bufferedoutputstream (system. out), true); n = SC. nextint (); For (INT I = 0; I <n; I ++) {M = SC. n Extint (); boo = false; Mark = new Boolean [m]; A = new integer [m]; sum = 0; edge = 0; For (Int J = 0; j <m; j ++) {A [J] = SC. nextint (); sum + = A [J];} arrays. sort (A, new comparator <integer> () {// sort public int compare (integer O1, integer O2) {return O1> O2? -; // Sort by descending order}); edge = sum> 2; // divide the right one by 2if (sum % 4 = 0 & edge> = A [0]) {DFS (0, 0); If (BOO) PW. println ("yes"); elsepw. println ("no");} else {PW. println ("no") ;}}// the first variable has several edges, and the second variable is the length of the edge of each loop, the third variable is the public static void DFS (INT length, int size, int index) element. {// if it is true, return if (BOO) return; // If (length = 4) {boo = true;} If (size = edge) {DFS (Length + 1, 0, length + 1);} else {for (INT I = index; I <m; I ++) {If (! Mark [I] & size + A [I] <= edge) {mark [I] = true; DFS (length, size + A [I], I + 1 ); mark [I] = false; // backtracking }}}}}

Method 2

Import Java. io. *; import Java. util. *; public class main {public static int n, m, sum, edge; public static Boolean mark []; public static int A []; public static void main (string [] ARGs) {consumer SC = new consumer (New bufferedinputstream (system. in); printwriter PW = new printwriter (New bufferedoutputstream (system. out), true); n = SC. nextint (); For (INT I = 0; I <n; I ++) {M = SC. nextint (); Mark = new Boolean [m]; A = new int [m]; sum = 0; f Or (Int J = 0; j <m; j ++) {A [J] = SC. nextint (); sum + = A [J];} edge = sum> 2; // shifts one digit to the right by 2arrays. sort (a); // sort if (sum % 4 = 0 & edge> = A [. length-1]) {// The side length must be greater than or equal to the maximum value of the elements in the array if (DFS (0, 0) PW. println ("yes"); elsepw. println ("no");} elsepw. println ("no") ;}// the first variable has several edges, and the second variable is the length of the edge of each loop, the third variable is the public static Boolean DFS (INT length, int size, int index) element. {// exit the loop if (length = 4) if it is equal to four edges) return true; If (size = edge) {If (DFS (Length + 1, 0, Length + 1)) Return true; elsereturn false;} else {for (INT I = index; I <m; I ++) {If (! Mark [I] & size + A [I] <= edge) {mark [I] = true; If (DFS (length, size + A [I], I + 1) return true; Mark [I] = false ;}}return false ;}}