1739: Magic Ball Problem - Minimum Path Coverage

Topic:

Assuming there are n pillars, you will now place balls numbered 1, 2, 3,... In the N pillars in turn according to the following rules.(1) Play the ball only on the top of a post at a time.(2) In the same post, the sum of the numbers of any two adjacent balls is a perfect square number.Try to design an algorithm to calculate the maximum number of balls that can be placed on N pillars.For example, you can put up to 11 balls on four columns.Programming task: For a given n, calculate the maximum number of balls that can be placed on N pillars.

Ideas:

The initial number of points is 1, each time the number of points is added one, a new edge is added, and the minimum path coverage is calculated once to determine if the value is greater than n, then exit, not continue the cycle.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>

using namespace std;

const int maxn = 10010;
const int INF = 0x3f3f3f3f;



struct Edge {
    int from, to, cap, flow, flag;
};
struct Dinic {
    int s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void init() {
        edges.clear();
        for (int i = 0; i < maxn; i++) G[i].clear();
    }
    void addedge(int from, int to, int cap, int flag) {
        edges.push_back(Edge{from, to, cap, 0, flag});
        edges.push_back(Edge{to, from, 0, 0, flag});
        int 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();
            for (int i = 0; i < G[x].size(); 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;
        for (int &i = cur[x]; i < G[x].size(); 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;
    }
}ac;


vector<int> G[maxn];
int vis[maxn];

void dfs(int u) {
    if (vis[u]) return;
    vis[u] = true;
    printf("%d ", u);
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        dfs(v);
    }
}

int main() {
    int n;
    scanf("%d", &n);
    int s = 0, t = 10001;
    int ans = 1, maxf = 0;
    while (true) {
        for (int i = 1; i < ans; i++) {
            int x = sqrt(i+ans);
            if (x * x != i + ans) continue;
            ac.addedge(i, ans+5000, 1, 1);
        }
        ac.addedge(s, ans, 1, 0);
        ac.addedge(ans+5000, t, 1, 0);
        maxf += ac.maxflow(s, t);
        if (ans - maxf > n) break;
        ans++;
    }
    printf("%d\n", ans-1);
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < ac.edges.size(); i += 2) {
        if (ac.edges[i].flag == 1 && ac.edges[i].flow == 1) {
            int u = ac.edges[i].from, v = ac.edges[i].to;
            G[u].push_back(v-5000);
        }
    }
    for (int i = 1; i < ans; i++) {
        if (vis[i]) continue;
        dfs(i);
        printf("\n");
    }
    return 0;
}

 

Tags: Programming

Posted on Wed, 05 Feb 2020 08:11:53 -0800 by kujtim