hdu4982 Goffi and Squary Partition （DFS解法）

BestCoder Round #6 B

http://acm.hdu.edu.cn/showproblem.php?pid=4982

Goffi and Squary Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 308    Accepted Submission(s): 106

Problem Description

Recently, Goffi is interested in squary partition of integers.
A set \(X\) of \(k\) distinct positive integers is called squary partition of \(n\) if and only if it satisfies the following conditions: [ol]

• the sum of \(k\) positive integers is equal to \(n\)
• one of the subsets of \(X\) containing \(k - 1\) numbers sums up to a square of integer.[/ol] For example, a set {1, 5, 6, 10} is a squary partition of 22 because 1 + 5 + 6 + 10 = 22 and 1 + 5 + 10 = 16 = 4 × 4.
  Goffi wants to know, for some integers \(n\) and \(k\), whether there exists a squary partition of \(n\) to \(k\) distinct positive integers.

 

Input

Input contains multiple test cases (less than 10000). For each test case, there's one line containing two integers \(n\) and \(k\) (\(2 \le n \le 200000, 2 \le k \le 30\)).

 

Output

For each case, if there exists a squary partition of \(n\) to \(k\) distinct positive integers, output "YES" in a line. Otherwise, output "NO".

 

Sample Input

2 2
4 2
22 4

 

Sample Output

NO
YES
YES

 

Source

BestCoder Round #6

 

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题意：给出n和k，求k个不同的正整数，使其中k-1个数能组成平方数，k个数的和为n。有解输出YES，无解输出NO。

题解：从大到小枚举小于n的平方数，剪枝搜索是否有解。

先判一下是不是1~k这最小的k个数加起来都超过n，这种情况肯定不可能，真去搜一遍的话太费时间了，还是特判掉比较好。

if((+k)*k/>n)return ;

枚举平方数，进入搜索：

     t=sqrt(1.0*n);
     if(t*t==n)t--;
     for(i=t; i>=; i--) {
         int lost=i*i;
         Remove(n-lost);
         if(dfs(lost,,R[n]))return ;
         Resume(n-lost);
     }

↑（这个在函数里，return 1是有解。）

这题每个数只用一次，所以枚举一个数，下个数可以从这个数+1开始枚举。我使用了DancingLinks那种超碉的结构，不过其实只用一个used[]就好了……

比赛的时候这样就能过了，因为n大的时候很容易就找到解，dfs不耗多长时间。不过今天一看居然加强了数据，会TLE，所以我们来加个剪枝：

当枚举第x个数时，剩下(k-x+1)个数，枚举i为当前的数，如果i+ (i+1)+(i+2)+...+(i+k-x)这以i开头的最小的k-x+1个数都大于剩下的数了，就不用继续枚举i了。

等差数列求和，(i + i+(k-x))*(k-x+1)/2 > lost，搞一搞可以把 i 移出来，得到i的最大值。

DFS代码：

 bool dfs(const int &lost,const int &x,const int &now) {
     if(x>k) {
         if(lost==)
             return ;
         else return ;
     }
     if(lost<=)return ;
     if(x!=k) {
         int maxi=((lost+lost)/(k-x+)-(k-x))/;///剩下的大于i的最小的(k-x+1)个数加起来超的话就跳
         maxi=min(maxi,n-);
         for(int i=now; i<=maxi; i=R[i]) {
             Remove(i);
             if(dfs(lost-i,x+,R[i]))return ;
             Resume(i);
         }
     } else {
         if(L[R[lost]]==lost) {
             Remove(lost);
             if(dfs(,x+,R[lost]))return ;
             Resume(lost);
         }
     }
     return ;
 }

（这题的正解好像是各种情况都考虑一下，直接几个if就判出来了……我是不太懂）（还有官方题解说平方数只用枚举最大的那个就行，可是8 2这组数据，最大的平方数是4，然后就逗乐啊……虽然好像只有这一组数据有问题）

全代码：

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usll unsigned ll
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) prllf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)
 #define mp make_pair
 #define pb push_back

 const int maxn=;
 int L[maxn],R[maxn];
 inline void Remove(int x) {
     L[R[x]]=L[x];
     R[L[x]]=R[x];
 }

 inline void Resume(int x) {
     L[R[x]]=x;
     R[L[x]]=x;
 }

 int n,k;

 bool dfs(const int &lost,const int &x,const int &now) {
     if(x>k) {
         if(lost==)
             return ;
         else return ;
     }
     if(lost<=)return ;
     if(x!=k) {
         int maxi=((lost+lost)/(k-x+)-(k-x))/;///剩下的大于i的最小的(k-x+1)个数加起来超的话就跳
         maxi=min(maxi,n-);
         for(int i=now; i<=maxi; i=R[i]) {
             Remove(i);
             if(dfs(lost-i,x+,R[i]))return ;
             Resume(i);
         }
     } else {
         if(L[R[lost]]==lost) {
             Remove(lost);
             if(dfs(,x+,R[lost]))return ;
             Resume(lost);
         }
     }
     return ;
 }

 bool farm(int _n,int _k) {
     int t;
     int i,j;
     n=_n;
     k=_k;
     if((+k)*k/>n)return ;
     for(i=; i<=n; i++) {
         L[i]=i-;
         R[i-]=i;
     }
     L[]=n;
     R[n]=;
     t=sqrt(1.0*n);
     if(t*t==n)t--;
     for(i=t; i>=; i--) {
         int lost=i*i;
         Remove(n-lost);
         if(dfs(lost,,R[n]))return ;
         Resume(n-lost);
     }
     return ;
 }

 int main() {
     int i;
     int n,k;
     while(scanf("%d%d",&n,&k)!=EOF) {
         if(farm(n,k))printf("YES\n");
         else printf("NO\n");
 //        for(i=0;i<r;i++)
 //            printf("%d ",b[i]);
 //        puts("");
     }
     return ;
 }