Number BZOJ3275 最大流

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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#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<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> 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() {
	queue<int>q;
	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
posted @ 2018-12-12 19:43  NKDEWSM  阅读(169)  评论(0编辑  收藏  举报