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

Sample Output22


Input

第一行一个正整数n，表示数的个数。n<=3000 第二行n个正整数a1,a2,...an Output

最大的和

Sample Input5 3 4 5 6 7  

首先可以发现，只有奇数和偶数才可能满足上面两个条件；那么我们把序列分为奇偶两部分；设源点st,汇点ed;st 和奇数连边，ed 和偶数连边；那么我们遍历可能的连边，O(n^2);这个连边权值设为 inf;(有点最大权闭合子图的意思？那么我们求最小割即可；最后就是sum-dinic()；

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;
inline ll rd() {
  ll x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}

ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
  ll ans = 1;
  a = a % c;
  while (b) {
    if (b % 2)ans = ans * a%c;
    b /= 2; a = a * a%c;
  }
  return ans;
}

int n, m;
int st, ed;
struct node {
  int u, v, nxt, w;
}edge[maxn<<2];

int head[maxn], cnt;
void addedge(int u, int v, int w) {
  edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w;
  edge[cnt].nxt = head[u]; head[u] = cnt++;
}
int rk[maxn];

int bfs() {
  queueq;
  ms(rk);
  rk[st] = 1; q.push(st);
  while (!q.empty()) {
    int tmp = q.front(); q.pop();
    for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
      int to = edge[i].v;
      if (rk[to] || edge[i].w <= 0)continue;
      rk[to] = rk[tmp] + 1; q.push(to);
    }
  }
  return rk[ed];
}

int dfs(int u, int flow) {
  if (u == ed)return flow;
  int add = 0;
  for (int i = head[u]; i != -1; i = edge[i].nxt) {
    int v = edge[i].v;
    if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
    int tmpadd = dfs(v, min(edge[i].w, flow - add));
    if (!tmpadd) { rk[v] = -1; continue; }
    edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd;
  }
  return add;
}

int ans;
void dinic() {
  while (bfs())ans += dfs(st, inf);
}

int a[maxn];

bool check(int a, int b) {
  if(a % 2 == 1 && b % 2 == 1)return 0;
  if (gcd(a, b) != 1)return 0;
  ll sum = a * a + b * b; int p = (int)sqrt(sum);
  if (p*p != sum)return 0;
  return 1;
}

int main()
{
  //ios::sync_with_stdio(0);
  memset(head, -1, sizeof(head));
  rdint(n);int sum = 0;
  for (int i = 1; i <= n; i++)rdint(a[i]), sum += a[i];
  st = n + 1; ed = st + 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)addedge(st, i, a[i]), addedge(i, st, 0);
    else addedge(i, ed, a[i]), addedge(ed, i, 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      if (check(a[i], a[j])) {
        if (a[i] & 1)addedge(i, j, inf), addedge(j, i, 0);
        else addedge(j, i, inf), addedge(i, j, 0);
      }
    }
  }
  dinic();
  cout << sum - ans << endl;
  return 0;
}

 

EPFL - Fighting