素数伴侣HJ28

题目链接：素数伴侣_牛客题霸_牛客网

解法：最大匹配

        因为素数不可能是偶数，所以“素数伴侣”只能是一个奇数和一个偶数。由此我们可以创建二分图：一个子集全是奇数，另一个子集全是偶数，若两个节点值之和为素数则连接一条边。二分图创建好后就是求最大匹配问题了，可以用匈牙利算法或者最大流算法求解（判断素数可以用线性筛）

代码示例1(匈牙利算法)：

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <type_traits>
#include <vector>
using namespace std;

const int V = 60010, N = 105, M = 5010;
vector<int> primes;
bool tag[V];

int h[N], to[M], ne[M], idx;
void Add(int u, int v) {
	to[idx] = v, ne[idx] = h[u], h[u] = idx++;
}

void Init() {
	fill_n(h, N, -1);

	primes.reserve(3000);
	for (int i = 2; i < V; i++) {
		if (!tag[i]) primes.push_back(i);
		for (auto j : primes) {
			if (i * j >= V) break;
			tag[i * j] = true;
			if (i % j == 0) break;
		}
	}
}

int ar[N];
int match[N];
bool flag[N];
int n;

bool DFS(int u) {
	flag[u] = true;
	for (int i = h[u]; ~i; i = ne[i]) {
		int v = to[i];
		if (flag[v]) continue;
		flag[v] = true;
		if (!match[v] || DFS(match[v])) {
			match[v] = u;
			return true;
		}
	}
	return false;
}

int main() {
	Init();
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", ar + i);
		for (int j = i - 1; j > 0; j--) {
			if (!tag[ar[i] + ar[j]]) {
				Add(i, j);
				Add(j, i);
			}
		}
	}

	int ans = 0;
	for (int i = 1; i <= n; i++) {
		if (ar[i] % 2) {
			fill_n(flag, N, false);
			if (DFS(i)) ++ans;
		}
	}

	printf("%d", ans);
}

代码示例2(最大流ISAP)

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <type_traits>
#include <vector>
#include <queue>
using namespace std;

const int V = 60010, N = 105, M = 5010;
vector<int> primes;
bool tag[V];

int h[N], to[M], ne[M], wt[M], idx;
void Add(int u, int v, int w) {
	to[idx] = v, wt[idx] = w, ne[idx] = h[u], h[u] = idx++;
}

void Init() {
	fill_n(h, N, -1);

	primes.reserve(3000);
	for (int i = 2; i < V; i++) {
		if (!tag[i]) primes.push_back(i);
		for (auto j : primes) {
			if (i * j >= V) break;
			tag[i * j] = true;
			if (i % j == 0) break;
		}
	}
}

int ar[N];
int n;
// 源点、汇点
int ss, tt;
// 各个结点所在层
int layer[N];
// 各层节点数量
int cnt[N];

void BFS() {
	queue<int> qu;
	qu.push(tt);
	layer[tt] = 1;
	cnt[1] = 1;
	int u, v;
	while (qu.size()) {
		u = qu.front();
		qu.pop();
		for (int i = h[u]; ~i; i = ne[i]) {
			v = to[i];
			if (layer[v]) continue;
			layer[v] = layer[u] + 1;
			++cnt[layer[v]];
			qu.push(v);
		}
	}
}

int DFS(int u, int from, int inflow) {
	if (u == tt) return inflow;
	int res = 0;
	for (int i = h[u]; ~i; i = ne[i]) {
		int v = to[i], & w = wt[i];
		if (layer[v] != layer[u] - 1 || !w) continue;
		int outflow = DFS(v, i, min(inflow, w));
		w -= outflow;
		if (~from) wt[from] += outflow;
		res += outflow;
		inflow -= outflow;
		if (!inflow) return res;
	}

	if (!--cnt[layer[u]++]) layer[ss] = n << 1;
	++cnt[layer[u]];
	return res;
}

int main() {
	Init();
	scanf("%d", &n);
	ss = n + 1, tt = n + 2;
	for (int i = 1; i <= n; i++) {
		scanf("%d", ar + i);
		if (ar[i] % 2) {
			Add(ss, i, 1);
			Add(i, ss, 0);
		}
		else {
			Add(i, tt, 1);
			Add(tt, i, 0);
		}
		for (int j = i - 1; j > 0; j--) {
			if (!tag[ar[i] + ar[j]]) {
				Add(i, j, 1);
				Add(j, i, 1);
			}
		}
	}

	int ans = 0;
	BFS();
	while (layer[ss] <= n + 2) ans += DFS(ss, -1, 1 << 30);

	printf("%d", ans);
}