luogu2765 魔术球问题 网络流

题目大意：

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。

 

关键字：网络流 拆点 上下界 二分查找

 

网络流：

想象拿柱子去串球，一个柱子就相当于一个流量为1的流。

 

拆点建图：

为保证每个球都能串在一个柱子上，将每个球拆成一条边，要求边内流量必须为1（即该边上下界都为1）。

若两球数字和为完全平方数，则将一球边的to节点连另一球边的from节点（注意球是按顺序放的，因此若a数字小于b数字，要么a连b，要么b连a，不能双向都连）。

为了满足网络流形式，“s”点向每个球边from点连边，每个球边to点向t点连边，容量为1。

因为柱子数量有限，因此将s点向“s”点连容量为柱子数量的边。

 

上下界：

根据上下界网络流处理办法，S点连每个球边to节点，每个球边from节点连T节点。之前的球边容量变为0，删去。t点向s点连容量∞边，即t，s点合并（合并点在代码中为orgT，"s"点为orgS）。

 

枚举：二分枚举球的数量直到得出答案。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cassert>
using namespace std;

//#define test
#define INF 0x3f3f3f3f
#define P(x) x*2+3
#define Q(x) x*2+4
#define arcP(x) (x-3)/2
#define arcQ(x) (x-4)/2
const int MAX_BALL = 2510, MAX_NODE = MAX_BALL * 3, MAX_EDGE = 50000, TOT_BALL = 2500;

struct Dinic
{
    struct Edge;
    struct Node;

    struct Node
    {
        int Id, Level;
        Edge* Head;
        Edge *DfsFrom;
        bool Deled;
        Node() { Deled = false; }
    };

    struct Edge
    {
        int Cap, OrgCap;
        Node *From, *To;
        Edge *Next, *Rev;
        Edge(int cap, Node *from, Node *to, Edge *next) :Cap(cap), OrgCap(cap), From(from), To(to), Next(next) {}
    };

    Node _nodes[MAX_NODE];
    Edge *_edges[MAX_EDGE];
    int _vCount, _eCount;
    Node *StartNode, *TargetNode;

    void Init(int sId, int tId, int vCount)
    {
        memset(_nodes, 0, sizeof(_nodes));
        memset(_edges, 0, sizeof(_edges));
        _eCount = 0;
        _vCount = vCount;
        StartNode = sId + _nodes;
        TargetNode = tId + _nodes;
    }

    void Reuse(Node *cur)
    {
        cur->Deled = false;
        for (Edge *e = cur->Head; e; e = e->Next)
        {
            if (!e->To->Deled)
            {
                e->Cap = e->OrgCap;
                e->Rev->Cap = e->Rev->OrgCap;
            }
        }
    }

    void ReuseById(int id)
    {
        Reuse(id + _nodes);
    }

    Edge* AddEdge(Node *from, Node *to, int eCap)
    {
        assert(from != to);
        Edge *cur = _edges[++_eCount] = new Edge(eCap, from, to, from->Head);
        cur->From->Head = cur;
        return cur;
    }

    void DeleteNode(Node *cur)
    {
        cur->Deled = true;
        for (Edge *e = cur->Head; e; e = e->Next)
        {
            e->Cap = e->Rev->Cap = 0;
        }
    }

    void DeleteById(int id)
    {
        DeleteNode(id + _nodes);
    }

    void Build(int uId, int vId, int eCap)
    {
        Node *u = uId + _nodes, *v = vId + _nodes;
        u->Id = uId;
        v->Id = vId;
        Edge *edge1 = AddEdge(u, v, eCap), *edge2 = AddEdge(v, u, 0);
        edge1->Rev = edge2;
        edge2->Rev = edge1;
    }

    struct NodeQueue
    {
        Node *q[MAX_NODE];
        int head, tail;
        void clear() { head = tail = 0; }
        void push(Node *v) { q[tail++] = v; }
        Node* front() { return q[head]; }
        void pop() { head++; }
        bool empty() { return head == tail; }
    };

    bool Bfs()//常规，从源点构造
    {
        for (int i = 1; i <= _vCount; i++)
            _nodes[i].Level = 0;
        static NodeQueue q;
        q.clear();
        StartNode->Level = 1;
        q.push(StartNode);
        while (!q.empty())
        {
            Node *u = q.front();
            q.pop();
            for (Edge *e = u->Head; e; e = e->Next)
            {
                assert(e->Cap >= 0);
                if (!e->To->Level && e->Cap)
                {
                    e->To->Level = u->Level + 1;
                    q.push(e->To);
                }
            }
        }
        return TargetNode->Level;//遗忘点
    }

    int Dfs(Node *cur, int limit)
    {
        if (cur == TargetNode)
            return limit;
        if (limit == 0)
            return 0;
        int curTake = 0;
        for (Edge *e = cur->DfsFrom; e; cur->DfsFrom = e = e->Next)
        {
            if (e->To->Level == cur->Level + 1 && e->Cap)
            {
                int nextTake = Dfs(e->To, min(limit - curTake, e->Cap));
                e->Cap -= nextTake;
                e->Rev->Cap += nextTake;
                curTake += nextTake;
            }
            if (limit - curTake == 0)
                break;
            //assert(e->Cap == 0);
        }
        return curTake;
    }


    int Proceed()
    {
        int ans = 0;
        while (Bfs())
        {
            for (int i = 0; i <= _vCount; i++)
                _nodes[i].DfsFrom = _nodes[i].Head;
            ans += Dfs(StartNode, INF);
#ifdef test
            printf("ans=%d\n", ans);
#endif
        }
        //printf("ans %d\n", ans);
        return ans;
    }
}g;
Dinic::Edge *ts;

bool IsSquare(int x)
{
    int a = floor(sqrt(x) + 0.5);
    return a*a == x;
}

bool Judge(int ballCnt)
{
    //printf("ballCnt %d\n", ballCnt);
    g._vCount = ballCnt * 2 + 4;
    ts->Cap = ts->OrgCap;
    ts->Rev->Cap = ts->Rev->OrgCap;
    for (int ball = ballCnt + 1; ball <= TOT_BALL; ball++)
    {
        g.DeleteById(P(ball));
        g.DeleteById(Q(ball));
    }
    for (int ball = 1; ball <= ballCnt; ball++)
    {
        g.ReuseById(P(ball));
        g.ReuseById(Q(ball));
    }
    return g.Proceed() == ballCnt;
}

int Bsearch(int maxL, int maxR)
{
    int l = maxL, r = maxR, ans = -1;
    while (l <= r)
    {
        int mid = (l + r) / 2;
        if (Judge(mid))
            l = (ans = mid) + 1;
        else
            r = mid - 1;
    }
    return ans;
}

void InitBuild(int totPole)
{
    int sId = 1, tId = 2, orgS = 3, orgT = 4;
    g.Init(sId, tId, 4);
    g.Build(orgT, orgS, totPole);
    ts = g._edges[g._eCount - 1];
    for (int ballNum = 1; ballNum <= TOT_BALL; ballNum++)
    {
        g._vCount += 2;
        g.Build(sId, Q(ballNum), 1);
        g.Build(P(ballNum), tId, 1);
        g.Build(orgS, P(ballNum), 1);
        g.Build(Q(ballNum), orgT, 1);
        for (int i = 1; i < ballNum; i++)
            if (IsSquare(i + ballNum))
                g.Build(Q(i), P(ballNum), 1);
    }
}

void Print(int ballNum)
{
    static bool Vis[MAX_BALL];
    for (int i = 1; i <= ballNum; i++)
    {
        if (Vis[i])
            continue;
        Dinic::Node *p = P(i) + g._nodes, *q = Q(i) + g._nodes;
        bool stop = false;
        while (!stop)
        {
            Vis[arcP(p->Id)] = true;
            printf("%d ", arcP(p->Id));
            stop = true;
            for (Dinic::Edge *e = q->Head; e; e = e->Next)
            {
                if (arcP(e->To->Id) <= ballNum)
                    //printf("print %d' - %d cap %d\n", arcQ(q->Id), arcP(e->To->Id), e->Cap);
                    if (!e->Cap && 1 <= arcP(e->To->Id) && arcP(e->To->Id) <= ballNum && !Vis[arcP(e->To->Id)] && !e->To->Deled)
                    {
                        stop = false;
                        p = e->To;
                        q = Q(arcP(p->Id)) + g._nodes;
                        break;
                    }
            }
        }
        printf("\n");
    }
}

int main()
{
#ifdef _DEBUG
    freopen("c:\\noi\\source\\input.txt", "r", stdin);
#endif
    int totPole;
    scanf("%d", &totPole);
    InitBuild(totPole);
    int ballNum = Bsearch(1, TOT_BALL);
    printf("%d\n", ballNum);
    Judge(ballNum);
    Print(ballNum);
    return 0;
}