网络流24题- 魔术球问题

网络流24题- 魔术球问题

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魔术球问题链接

Description 
    假设有n根柱子，现要按下述规则在这n根柱子中依次放入编号为1，2，3，...的球。
    （1）每次只能在某根柱子的最上面放球。
    （2）在同一根柱子中，任何2个相邻球的编号之和为完全平方数。
    试设计一个算法，计算出在n根柱子上最多能放多少个球。例如，在4 根柱子上最多可
    放11 个球。
    编程任务：
    对于给定的n，计算在n根柱子上最多能放多少个球。
Input 
    4
Output 
    11
    1 8
    2 7 9
    3 6 10
    4 5 11

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我们把x, y 满足(x < y && x + y是完全平方数)连一条边，我们要用尽量少的柱子放尽可能多的边。即用最少的路径覆盖尽可能多的点。问题就转化成了求有向图的最小路径覆盖。那么我们要求的就是如果1~n的数进行连边，求得的最小路径覆盖大于我们的柱子数。那么我们的答案就是对1~n-1的点求最小路径覆盖。并求出最小路径覆盖的路径

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首先是一发及其暴力的枚举dinic，打个表，提前算出55个答案是在数字几的dinic的图中得出了，这个代码暴力TLE但是如果打出了表那么复杂度就是一次dinic的

#include <bits/stdc++.h>

using namespace std;

int k;

struct Dinic {

    static const int MAXN = 30005 + 7;
    static const int MAXM = 1e7 + 7;
    static const int INF = 0x3f3f3f3f;

    int n, m, s, t;

    int first[MAXN], cur[MAXN], dist[MAXN], sign;

    struct Node {
        int to, flow, next;
    } edge[MAXM];

    inline void init(int start, int vertex, int ss, int tt) {
        n = vertex, s = ss, t = tt;
        for(int i = start; i <= n; i++ ) {
            first[i] = -1;
        }
        sign = 0;
    }

    inline void addEdge(int u, int v, int flow) {
        edge[sign].to = v, edge[sign].flow = flow, edge[sign].next = first[u];
        first[u] = sign++;
    }

    inline void add_edge(int u, int v, int flow) {
        addEdge(u, v, flow);
        addEdge(v, u, 0);
    }

    int match[MAXN] = {0}, vis[MAXN] = {0};

    void show_graph() {
        for(int i = 1; i <= k; i++ ) {
            for(int j = first[i]; ~j; j = edge[j].next) {
                int to = edge[j].to, w = edge[j].flow;
                if(to != s && edge[j ^ 1].flow) {
                    //printf("%d ", to - k);
                    match[i] = to - k;
                }
            }
        }
//        for(int i = 1; i <= k; i++ ) {
//            printf("[%d %d]
", i, match[i]);
//        }
        for(int i = 1; i <= k; i++ ) {
            int now = i;
            if(vis[i]) {
                continue;
            }
            while(1) {
                printf("%d", now);
                vis[now] = 1;
                now = match[now];
                if(now == 0) {
                    puts("");
                    break;
                } else {
                    printf(" ");
                }
            }
        }
    }

    inline int dinic() {
        int max_flow = 0;
        while(bfs(s, t)) {
            for(int i = 0; i <= n; i++ ) {
                cur[i] = first[i];
            }
            max_flow += dfs(s, INF);
        }
        return max_flow;
    }

    bool bfs(int s, int t) {
        memset(dist, -1, sizeof(dist));
        queue<int>que;
        que.push(s), dist[s] = 0;
        while(!que.empty()) {
            int now = que.front();
            que.pop();
            if(now == t) {
                return 1;
            }
            for(int i = first[now]; ~i; i = edge[i].next) {
                int to = edge[i].to, flow = edge[i].flow;
                if(dist[to] == -1 && flow > 0) {
                    dist[to] = dist[now] + 1;
                    que.push(to);
                }
            }
        }
        return 0;
    }

    int dfs(int now, int max_flow) {
        if(now == t) {
            return max_flow;
        }
        int ans = 0, next_flow = 0;
        for(int &i = cur[now]; ~i; i = edge[i].next) {
            int to = edge[i].to, flow = edge[i].flow;
            if(dist[to] == dist[now] + 1 && flow > 0) {
                next_flow = dfs(to, min(max_flow - ans, flow));
                ans += next_flow;
                edge[i].flow -= next_flow;
                edge[i ^ 1].flow += next_flow;
                if(ans == max_flow) {
                    return max_flow;
                }

            }
        }
        if(ans == 0) {
            return dist[now] = 0;
        }
        return ans;
    }

} cwl;

int main() {
    int n;
    scanf("%d", &n);
    for(k = 1; k; k++ ) {
        cwl.init(0, 2 * k + 1, 0, 2 * k + 1);
        for(int i = 1; i <= k; i++ ) {
            cwl.add_edge(0, i, 1);
            cwl.add_edge(i + k, 2 * k + 1, 1);
        }
        for(int i = 1; i <= k; i++ ) {
            for(int j = i + 1; j <= k; j++ ) {
                int val = i + j;
                int sqr = sqrt(val);
                if(sqr * sqr == val) {
                    cwl.add_edge(i, j + k, 1);
                }
            }
        }
        int ans = cwl.dinic();
        if(k - ans > n) {
            printf("%d
", k - 1);
            k--;
            cwl.init(0, 2 * k + 1, 0, 2 * k + 1);
            for(int i = 1; i <= k; i++ ) {
                cwl.add_edge(0, i, 1);
                cwl.add_edge(i + k, 2 * k + 1, 1);
            }
            for(int i = 1; i <= k; i++ ) {
                for(int j = i + 1; j <= k; j++ ) {
                    int val = i + j;
                    int sqr = (int)sqrt(val);
                    if(sqr * sqr == val) {
                        cwl.add_edge(i, j + k, 1);
                    }
                }
            }
            ans = cwl.dinic();
            cwl.show_graph();
            break;
        }
    }

    return 0;
}

然后是用打的表之间知道答案一次dinic求答案，24MS。

#include <bits/stdc++.h>

using namespace std;

int k;

int ans[] = {0, 1,3,7,11,17,23,31,39,49,59,71,83,97,111,127,143,161,179,199,219,241,263,287,311,337,363,391,419,449,479,511,543,577,611,647,683,721,759,799,839,881,923,967,1011,1057,1103,1151,1199,1249,1299,1351,1403,1457,1511,1567};

struct Dinic {

    static const int MAXN = 30005 + 7;
    static const int MAXM = 1e7 + 7;
    static const int INF = 0x3f3f3f3f;

    int n, m, s, t;

    int first[MAXN], cur[MAXN], dist[MAXN], sign;

    struct Node {
        int to, flow, next;
    } edge[MAXM];

    inline void init(int start, int vertex, int ss, int tt) {
        n = vertex, s = ss, t = tt;
        for(int i = start; i <= n; i++ ) {
            first[i] = -1;
        }
        sign = 0;
    }

    inline void addEdge(int u, int v, int flow) {
        edge[sign].to = v, edge[sign].flow = flow, edge[sign].next = first[u];
        first[u] = sign++;
    }

    inline void add_edge(int u, int v, int flow) {
        addEdge(u, v, flow);
        addEdge(v, u, 0);
    }

    int match[MAXN] = {0}, vis[MAXN] = {0};

    void show_graph() {
        for(int i = 1; i <= k; i++ ) {
            for(int j = first[i]; ~j; j = edge[j].next) {
                int to = edge[j].to, w = edge[j].flow;
                if(to != s && edge[j ^ 1].flow) {
                    //printf("%d ", to - k);
                    match[i] = to - k;
                }
            }
        }
//        for(int i = 1; i <= k; i++ ) {
//            printf("[%d %d]
", i, match[i]);
//        }
        for(int i = 1; i <= k; i++ ) {
            int now = i;
            if(vis[i]) {
                continue;
            }
            while(1) {
                printf("%d", now);
                vis[now] = 1;
                now = match[now];
                if(now == 0) {
                    puts("");
                    break;
                } else {
                    printf(" ");
                }
            }
        }
    }

    inline int dinic() {
        int max_flow = 0;
        while(bfs(s, t)) {
            for(int i = 0; i <= n; i++ ) {
                cur[i] = first[i];
            }
            max_flow += dfs(s, INF);
        }
        return max_flow;
    }

    bool bfs(int s, int t) {
        memset(dist, -1, sizeof(dist));
        queue<int>que;
        que.push(s), dist[s] = 0;
        while(!que.empty()) {
            int now = que.front();
            que.pop();
            if(now == t) {
                return 1;
            }
            for(int i = first[now]; ~i; i = edge[i].next) {
                int to = edge[i].to, flow = edge[i].flow;
                if(dist[to] == -1 && flow > 0) {
                    dist[to] = dist[now] + 1;
                    que.push(to);
                }
            }
        }
        return 0;
    }

    int dfs(int now, int max_flow) {
        if(now == t) {
            return max_flow;
        }
        int ans = 0, next_flow = 0;
        for(int &i = cur[now]; ~i; i = edge[i].next) {
            int to = edge[i].to, flow = edge[i].flow;
            if(dist[to] == dist[now] + 1 && flow > 0) {
                next_flow = dfs(to, min(max_flow - ans, flow));
                ans += next_flow;
                edge[i].flow -= next_flow;
                edge[i ^ 1].flow += next_flow;
                if(ans == max_flow) {
                    return max_flow;
                }

            }
        }
        if(ans == 0) {
            return dist[now] = 0;
        }
        return ans;
    }

} cwl;

int get_value(int n) {
    printf("%d
", n);
    k = n;
    cwl.init(0, 2 * k + 1, 0, 2 * k + 1);
    for(int i = 1; i <= k; i++ ) {
        cwl.add_edge(0, i, 1);
        cwl.add_edge(i + k, 2 * k + 1, 1);
    }
    for(int i = 1; i <= k; i++ ) {
        for(int j = i + 1; j <= k; j++ ) {
            int val = i + j;
            int sqr = sqrt(val);
            if(sqr * sqr == val) {
                cwl.add_edge(i, j + k, 1);
            }
        }
    }
    int ans = cwl.dinic();
    cwl.show_graph();
}

int main() {
    int n;
    scanf("%d", &n);
    get_value(ans[n]);
    return 0;
}