【HDU - 1518】Square （经典的dfs + 剪枝）

题干：

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

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

Sample Output

yes
no
yes

题目大意：

给定一些长度各异的木棒，请问是否可能把它们连接成一个正方形？

解题报告：

   这题是非常经典的dfs剪枝问题，但是一直弄不懂为什么必须要搞成   “一次dfs凑三条边  ”才可以，不能把dfs的功能限制成 凑一条边，然后进行三次dfs吗？（反正我是wa了。。。逃）

ps：与这道经典的剪枝题 相见恨晚啊！！

AC代码：

#include

using namespace std;
int n,sum;
int a[55];
bool vis[55];

bool dfs(int cur, int st,int res) {
	if(cur == 4) return 1;
	if(res < 0) return 0 ;
	if(res == 0) {
		int bg = 1;
		while(vis[bg]) ++bg;
		for(int i = bg; i<=n; i++) {
			if(vis[i]) continue;
			vis[i]=1; 
			if(dfs(cur+1,i+1,sum/4-a[i])) return 1;
			vis[i]=0;
		}
		return 0;
	}
//	if(st > n) return 0;
	for(int i = st; i<=n; i++) {
		if(vis[i]) continue;
		vis[i]=1;
		if(dfs(cur,i+1,res - a[i])) return 1;
		vis[i]=0;
	}
	return 0;
}
bool cmp(const int & a,const int & b) {
	return a > b;
}
int main()
{
	int t;
	cin>>t;
	while(t--) {
		scanf("%d",&n);
		sum=0;
		int flag=1;
		for(int i = 1; i<=n; i++) scanf("%d",a+i),sum+=a[i]; 
		if(sum%4 != 0 ){
			printf("no\n");continue;
		}
		memset(vis,0,sizeof vis);
		sort(a+1,a+n+1,cmp);
		if(dfs(1,1,sum/4)) printf("yes\n");
		else printf("no\n");
	}
	return 0 ;
}

wa代码：

#include

using namespace std;
int n,sum;
int a[55];
bool vis[55];

bool dfs(int st,int res) {
	if(res < 0) return 0 ;
	if(res == 0) return 1 ;
	for(int i = st; i<=n; i++) {
		if(vis[i]) continue;
		vis[i]=1;
		if(dfs(i+1,res - a[i])) return 1;
		vis[i]=0;
	}
	return 0;
}
bool cmp(const int & a,const int & b) {
	return a > b;
}
int main()
{
	int t;
	cin>>t;
	while(t--) {
		scanf("%d",&n);
		sum=0;
		int flag=1;
		for(int i = 1; i<=n; i++) scanf("%d",a+i),sum+=a[i]; 
		if(sum%4 != 0 ){
			printf("no\n");continue;
		}
		memset(vis,0,sizeof vis);
		sort(a+1,a+n+1,cmp);
		int st = 1;
		for(int i = 1; i<=3; i++) {
			while(vis[st]) ++st;
			if(!dfs(st,sum/4)) {
				flag=0;break;
			}	
		}
		if(flag) printf("yes\n");
		else printf("no\n");
	}
	return 0 ;
}

ps：

求大神给个解释，，至今不知道为什么、、、