BZOJ 3275: Number

3275: Number

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 874  Solved: 371
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Description

有N个正整数，需要从中选出一些数，使这些数的和最大。
若两个数a,b同时满足以下条件，则a,b不能同时被选
1:存在正整数C，使a*a+b*b=c*c
2:gcd(a,b)=1

Input

第一行一个正整数n，表示数的个数。

第二行n个正整数a1,a2,?an。

 

 

Output

最大的和。

 

Sample Input

5
3 4 5 6 7

Sample Output

22

HINT

n<=3000。

Source

网络流

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机制建图，同 BZOJ 3158: 千钧一发 。

 #include <cmath>
 #include <cstdio>
 #include <cstring>

 typedef long long lnt;

 const int siz = ;
 const int inf = ;

 int n;
 int a[siz];
 int b[siz];

 int tot;
 int s, t;
 int hd[siz];
 int to[siz];
 int fl[siz];
 int nt[siz];

 inline void add(int u, int v, int f)
 {
     nt[tot] = hd[u]; to[tot] = v; fl[tot] = f; hd[u] = tot++;
     nt[tot] = hd[v]; to[tot] = u; fl[tot] = ; hd[v] = tot++;
 }

 int dep[siz];

 inline bool bfs(void)
 {
     static int que[siz];
     static int head, tail;

     memset(dep, , sizeof(dep));
     head = , tail = ;
     que[tail++] = s;
     dep[s] = ;

     while (head != tail)
     {
         int u = que[head++], v;

         for (int i = hd[u]; ~i; i = nt[i])
             if (!dep[v = to[i]] && fl[i])
                 dep[que[tail++] = v] = dep[u] + ;
     }

     return dep[t];
 }

 int cur[siz];

 int min(int a, int b)
 {
     return a < b ? a : b;
 }

 int dfs(int u, int f)
 {
     if (u == t || !f)
         return f;

     int used = , flow, v;

     for (int i = cur[u]; ~i; i = nt[i])
         if (dep[v = to[i]] == dep[u] +  && fl[i])
         {
             flow = dfs(v, min(f - used, fl[i]));

             used += flow;
             fl[i] -= flow;
             fl[i^] += flow;

             if (used == f)
                 return f;

             if (fl[i])
                 cur[u] = i;
         }

     if (!used)
         dep[u] = ;

     return used;
 }

 inline int maxFlow(void)
 {
     int maxFlow = , newFlow;

     while (bfs())
     {
         for (int i = s; i <= t; ++i)
             cur[i] = hd[i];

         while (newFlow = dfs(s, inf))
             maxFlow += newFlow;
     }

     return maxFlow;
 }

 inline lnt sqr(lnt x)
 {
     return x * x;
 }

 int gcd(int x, int y)
 {
     return y ? gcd(y, x % y) : x;
 }

 inline bool check(int x, int y)
 {
     if (gcd(x, y) != )
         return false;

     lnt t = sqr(x) + sqr(y);
     if (sqr(sqrt(t)) != t)
         return false;

     return true;
 }

 signed main(void)
 {
     scanf("%d", &n);

     for (int i = ; i <= n; ++i)
         scanf("%d", a + i);

     for (int i = ; i <= n; ++i)
         b[i] = a[i];

     s = , t = n + ;

     memset(hd, -, sizeof(hd));

     for (int i = ; i <= n; ++i)
         if (a[i] & )
             add(s, i, b[i]);
         else
             add(i, t, b[i]);

     for (int i = ; i <= n; ++i)
         for (int j = ; j <= n; ++j)
             if (a[i] & )if (!(a[j] & ))
                 if (check(a[i], a[j]))
                     add(i, j, inf);

     int sum = ;

     for (int i = ; i <= n; ++i)
         sum += b[i];

     printf("%d\n", sum - maxFlow());
 }

@Author: YouSiki