hdu1518 (Square) deep search + pruning

Click to open Hangzhou electric 1518

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 is begins with an integer 4 <= M <=, the number of sticks. M integers follow; Each gives the length of a stick-an integer between 1 and 10,000.

Outputfor each case, the output a line containing "yes" if was is the 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

Test instructions: Judging whether the given number of numbers can form a positive quadrilateral

Idea: From the test instructions can be seen, to become a positive quadrilateral, then must be given the sum of the number and can not be divided into 4, and the smallest number of the given number can not be greater than the side length of the quadrilateral. And then through the deep search through each number, until the length of the combination into a side length, and finally through the statistical composition of the number of edge length, determine whether the condition is satisfied

Import Java.util.scanner;public class P1518 {public static int m,edglen;public static int[] visit;public static int[] A;pu Blic static Boolean flag;public static void Main (string[] args) {Scanner sc=new Scanner (system.in); int n=sc.nextint (); whi Le (n-->0) {m=sc.nextint (); A=new int[m];visit=new int[m];int sum=0;flag=false;for (int i=0;i<m;i++) {a[i]= Sc.nextint (); sum+=a[i];} if (sum%4!=0) {//pruning, sum cannot be divisible by 4, you must not make up System.out.println ("no"); continue;} Edglen=sum/4;sort (); if (Edglen<a[0]) {//the smallest is greater than the length of the edge, minus System.out.println ("no"); continue;} DFS (0,a[0],0); if (flag) {System.out.println ("yes");} Else{system.out.println ("No");}}} private static void DFS (int edglennumb, int edg, int i) {int k=i;visit[i]=1;if (edg==edglen) {edglennumb++;k=0;edg=0;} if (edglennumb==3) {Flag=true;return;} for (int j=k;j<m;j++) {if (visit[j]==0 && edg+a[j]<=edglen) {DFS (edglennumb,a[j]+edg,j); if (flag) {return ;}}} visit[i]=0;} private static void sort () {for (int. i=0;i<m-1;i++) {for (int j=0;j<m-1-i;j++) {if (a[i]>a[j]) {int T=a[i];a[i]=a[j];a[j]=t;}}}} 

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hdu1518 (Square) deep search + pruning