【HDU - 1518】Square （经典的dfs + 剪枝）

原创

wx62a8776062513 2022-06-15 10:18:47 博主文章分类：dfs+剪枝 ©著作权

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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<bits/stdc++.h>

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<bits/stdc++.h>

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 ;
}