拆点 + 最大流

G - Erasing Prime Pairs (atcoder.jp)

题意

有 n（n <= 100）种互不相同的数，分别是 \(A[i]\) （<=1e7）, 每个有 \(B[i]\) 个

每次可以任意取两个数，如果相加是素数就消去这两个数，求最多操作次数

思路

思路一、

1. 不考虑 1 + 1 = 2 出现偶素数，可以将奇数，偶数分开，能消掉的连边，形成二分图，跑一下最大流即可
2. 考虑 1 + 1 = 2，可以设有 x 次 1 + 1 的操作，可以证明最多次数随 x 是单峰函数，三分即可

思路二、

​ 对于每个点都拆成入点和出点（解决了两个相同的数相加构成素数的情况），直接跑最大流，结果除以二即可

#include <bits/stdc++.h>
#define itn int
#define int long long
#define endl "\n"
#define PII pair<int, int>
using namespace std;
const int N = 1010;
const int M = 2e7 + 10;
const itn inf = 0x3f3f3f;
const int mod = 998244353;
// const int mod = 1e9 + 7;

int n, m, s, t;

struct Edge {
    int from, to, cap, flow;
    Edge(int f, int t, int c, int fl) {
        from = f;
        to = t;
        cap = c;
        flow = fl;
    }
};

struct Dinic {
    int n, m, s, t;  //结点数，边数（包括反向弧），源点编号和汇点编号
    vector<Edge> edges;  //边表。edge[e]和edge[e^1]互为反向弧
    vector<int> G[N];  //邻接表，G[i][j]表示节点i和第j条边在e数组中的序号
    bool vis[N];  // BFS使用
    int d[N];     //从起点到i的距离
    int cur[N];   //当前弧下标

    void clear_all(int n) {
        for (int i = 0; i < n; i++)
            G[i].clear();
        edges.clear();
    }
    void clear_flow() {
        int len = edges.size();
        for (int i = 0; i < len; i++)
            edges[i].flow = 0;
    }
    void add_edge(int from, int to, int cap) {
        edges.push_back(Edge(from, to, cap, 0));
        edges.push_back(Edge(to, from, 0, 0));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }
    bool BFS() {
        memset(vis, 0, sizeof(vis));
        queue<int> q;
        q.push(s);
        d[s] = 0;
        vis[s] = 1;
        while (!q.empty()) {
            int x = q.front();
            q.pop();
            int len = G[x].size();
            for (int i = 0; i < len; i++) {
                Edge& e = edges[G[x][i]];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int DFS(int x, int a) {
        if (x == t || a == 0)
            return a;
        int flow = 0, f, len = G[x].size();
        for (int& i = cur[x]; i < len; i++) {
            Edge& e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] &&
                (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[G[x][i] ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0)
                    break;
            }
        }
        return flow;
    }
    int maxflow(int s, int t) {
        this->s = s;
        this->t = t;
        int flow = 0;
        while (BFS()) {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, inf);
        }
        return flow;
    }

    int mincut() {  // call this after maxflow
        int ans = 0;
        int len = edges.size();
        for (int i = 0; i < len; i++) {
            Edge& e = edges[i];
            if (vis[e.from] && !vis[e.to] && e.cap > 0)
                ans++;
        }
        return ans;
    }
    void reduce() {
        int len = edges.size();
        for (int i = 0; i < len; i++)
            edges[i].cap -= edges[i].flow;
    }
} dinic;

bool vis[M], isprime[M];

void init() {
    for (int i = 2; i <= M; i++) {
        if (vis[i])
            continue; 
        isprime[i] = 1;
        for (int j = i + i; j <= M; j += i)
            vis[j] = 1;
    }
}

int a[N], b[N];

void solve() {
    itn n;
    cin >> n;
    init();
    int S = 0, T = 2 * n + 1;
    for (int i = 1; i <= n; i++) {
        cin >> a[i] >> b[i];
        dinic.add_edge(S, i, b[i]);
        dinic.add_edge(i + n, T, b[i]);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (isprime[a[i] + a[j]])
                dinic.add_edge(i, j + n, 1e18);
        }
    }
    cout << dinic.maxflow(S, T) / 2 << endl;
}

signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(12);
    solve();
    return 0;
}