题目
思路
看到配对，想到网络流。
考虑如果一个点是奇数，那么将源点与其连接，如果是偶数，那么将汇点与其连接，如果一对奇数和偶数的和是质数，那么将它们两对应的点相连。其中，我们要对 1 特殊处理，因为 \(1 + 1 = 2\) 而 \(2\) 是偶数且是质数，所以考虑费用流，尽可能多地保留 \(1\)，对所有不是 \(1\) 的奇数，连边不要费用，对于 \(1\)，费用为 \(1\)。最后答案为 \(maxflow + \left\lfloor\dfrac{mincost}{2}\right\rfloor\)。
代码
#include <bits/stdc++.h>#define int long longusing namespace std;const int N = 5010, M = 100010, INF = 0x3f3f3f3f3f3f3f3f;struct edge {int to, next, w, cost;
} e[M];int head[N], idx = 1;void add(int u, int v, int w, int cost) {idx++;e[idx].to = v;e[idx].next = head[u];e[idx].w = w;e[idx].cost = cost;head[u] = idx;idx++;e[idx].to = u;e[idx].next = head[v];e[idx].w = 0;e[idx].cost = -cost;head[v] = idx;
}int n, S, T;
int dis[N], pre[N], flow[N];
bool st[N];bool spfa() {queue<int> q;q.push(S);memset(dis, 0x3f, sizeof(dis));memset(flow, 0, sizeof(flow));dis[S] = 0;flow[S] = INF;st[S] = true;while (q.size()) {int t = q.front();q.pop();st[t] = false;for (int i = head[t]; i; i = e[i].next) {int to = e[i].to;if (e[i].w && dis[to] > dis[t] + e[i].cost) {dis[to] = dis[t] + e[i].cost;pre[to] = i;flow[to] = min(flow[t], e[i].w);if (!st[to]) {q.push(to);st[to] = true;}}}}return flow[T] > 0;
}int a[N], b[N];bool prime(int x) {if (x <= 2) return false;for (int i = 2; i <= x / i; i++) {if (x % i == 0) {return false;}}return true;
}signed main() {ios::sync_with_stdio(false);cin.tie(nullptr);cin >> n;for (int i = 1; i <= n; i++) cin >> a[i] >> b[i];S = n + 1, T = n + 2;int cnt1 = 0;for (int i = 1; i <= n; i++) {if (a[i] % 2 == 1) {add(S, i, b[i], a[i] == 1);if (a[i] == 1) cnt1 += b[i];}else {add(i, T, b[i], 0);}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {if (prime(a[i] + a[j]) && a[i] % 2 == 1 && a[j] % 2 == 0) {add(i, j, 0x3f3f3f3f3f3f3f3f, 0);}}}int maxflow = 0, mincost = 0;while (spfa()) {maxflow += flow[T];mincost += flow[T] * dis[T];int x = T;while (x != S) {e[pre[x]].w -= flow[T];e[pre[x] ^ 1].w += flow[T];x = e[pre[x] ^ 1].to;}}cout << maxflow + (cnt1 - mincost) / 2 << '\n';return 0;
}
