hdu 1518 Square dfs深搜算法

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16313    Accepted Submission(s): 5152

Problem Description

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

 

Input

The 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.

 

Output

For each case, output a line containing "yes" if is 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

 

Source

University of Waterloo Local Contest 2002.09.21

dfs函数中搜索的for循环中的剪枝是关键

//hdu1016//#include"stdafx.h"#include<iostream>#include<algorithm>using namespace std;#define size 30int map[size];bool visit[size];bool flag = false;int n, edgeLength;//index:搜索的下标 edgeCount:已经拼成的边的数目 preLength:已经拼成的长度void dfs(int index,int edgeCount,int preLength) {if (flag)return;if (edgeCount == 4) {flag = true;return;}//如果已经拼成的长度等于该正方形的边长 便你了下标从0开始if (preLength == edgeLength) {dfs(0, edgeCount + 1, 0);if (flag)return;}//for循环中的i=index是关键 减少遍历的次数 剪枝for (int i = index; i < n; i++) {if (visit[i])continue;int lengthNow = preLength + map[i];if (lengthNow > edgeLength)continue;visit[i] = true;dfs(i + 1, edgeCount, lengthNow);visit[i] = false;if (flag)return;}}int main() {int cases;cin >> cases;while (cases--) {int sum = 0;cin >> n;for (int i = 0; i < n; i++) {cin >> map[i];sum += map[i];}if (sum % 4 != 0) {goto No;}edgeLength = sum / 4;for (int i = 0; i < n; i++) {if (map[i] > edgeLength)goto No;}sort(map, map + n);flag = false;memset(visit, false, sizeof(visit));dfs(0,0,0);if (flag)cout << "yes" << endl;elseNo:cout << "no" << endl;}return 0;}