BZOJ3275 Number <黑白染色+最小割>

Problem

Number

$\mathrm{Time\;Limit:\;10\;Sec}$
$\mathrm{Memory\;Limit:\;128\;MB}$

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$a^2+b^2=c^2$
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$a_1,a_2,\cdots ,a_N$。

Output

最大的和。

Sample Input

5
3 4 5 6 7

Sample Output

22

HINT

$N\le 3000$。

标签：黑白染色 最小割

Solution

两条性质：
·奇$^2$+奇$^2$ $\ne$ $X^2 (X\in N^*)$
·$\gcd($偶,偶$) \ne 1$
∴奇与奇共存，偶与偶共存
建模：$S\to$奇 $ : val$，偶$\to T : val$，奇$\to$偶 $: INF$
$ans = $ 总权值$-$最小割

Code

#include <bits/stdc++.h>
#define MAX_N 3000
#define INF 2147483647
using namespace std;
int n, s, t, sum, cnt, a[MAX_N+5], d[MAX_N+5], pr[MAX_N+5];
struct node {int u, v, c, nxt;} E[MAX_N*100];
int gcd(int x, int y) {return y ? gcd(y, x%y) : x;}
void init() {s = 0, t = n+1; memset(pr, -1, sizeof pr);}
void insert(int u, int v, int c) {E[cnt].v = v, E[cnt].c = c, E[cnt].nxt = pr[u], pr[u] = cnt++;}
void addedge(int u, int v, int c) {insert(u, v, c), insert(v, u, 0);}
bool BFS() {
    queue <int> que;	que.push(s);
    memset(d, -1, sizeof d), d[s] = 0;
    while (!que.empty()) {
        int u = que.front();	que.pop();
        for (int i = pr[u]; ~i; i = E[i].nxt) {
            int v = E[i].v, c = E[i].c;
            if (!c || ~d[v])	continue;
            d[v] = d[u]+1, que.push(v);
        }
    }
    return ~d[t];
}
int DFS(int u, int flow) {
    if (u == t)	return flow;
    int ret = 0;
    for (int i = pr[u]; ~i; i = E[i].nxt) {
        int v = E[i].v, c = E[i].c;
        if (!c || d[v] != d[u]+1)	continue;
        int tmp = DFS(v, min(flow, c));
        E[i].c -= tmp, E[i^1].c += tmp;
        flow -= tmp, ret += tmp;
        if (!flow)	break;
    }
    if (!ret) d[u] = -1;
    return ret;
}
int Dinic() {int ret = 0;    while (BFS()) ret += DFS(s, INF);	return ret;}
bool chk(int i, int j) {return gcd(a[i], a[j]) == 1 && sqrt(a[i]*a[i]+a[j]*a[j]) == floor(sqrt(a[i]*a[i]+a[j]*a[j]));}
int main() {
    scanf("%d", &n), init();
    for (int i = 1; i <= n; i++) scanf("%d", a+i), sum += a[i];
    for (int i = 1; i <= n; i++)
        if (a[i]%2) addedge(s, i, a[i]);
        else addedge(i, t, a[i]);
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++)
        if (a[i]%2 == 1 && a[j]%2 == 0 && chk(i, j)) addedge(i, j, INF);
    return printf("%d", sum-Dinic()), 0;
}