链接:

题意:

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport.

It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

思路:

建图, 但是不能对每个商品同时建图,每个商品矩阵分别建图,同时不满足条件就不要跑费用流了..会T.

代码:

#include <iostream>

#include <cstdio>

#include <cstring>

#include <vector>

//#include <memory.h>

#include <queue>

#include <set>

#include <map>

#include <algorithm>

#include <math.h>

#include <stack>

#include <string>

#define MINF 0x3f3f3f3f

using namespace std;

typedef long long LL;

const int MAXN = 50+10;

const int INF = 1e9;

struct Edge

{

    int from, to, flow, cap, cost;

    Edge(int from, int to, int flow, int cap, int cost)

    {

        this->from = from;

        this->to = to;

        this->flow = flow;

        this->cap = cap;

        this->cost = cost;

    }

};

vector<Edge> edges;

vector<int> G[MAXN*MAXN*MAXN];

int Sh[MAXN][MAXN];

int Wo[MAXN][MAXN];

int SumN[MAXN];

int a[MAXN*MAXN];

int Vis[MAXN*MAXN*MAXN], Dis[MAXN*MAXN*MAXN], Pre[MAXN*MAXN*MAXN];

int n, m, k;

int s, t;

void AddEdge(int from, int to, int cap, int cost)

{

    edges.push_back(Edge(from, to, 0, cap, cost));

    edges.push_back(Edge(to, from, 0, 0, -cost));

    int len = edges.size();

    G[from].push_back(len-2);

    G[to].push_back(len-1);

}

bool SPFA()

{

    memset(Dis, MINF, sizeof(Dis));

    memset(Vis, 0, sizeof(Vis));

    queue<int> que;

    Dis[s] = 0;

    Vis[s] = 1;

    que.push(s);

    a[s] = INF;

    while (!que.empty())

    {

//        for (int i = s;i <= t;i++)

//            cout << Dis[i] << ' ' ;

//        cout << endl;

        int u = que.front();

//        cout << u << endl;

        que.pop();

        Vis[u] = 0;

        for (int i = 0;i < G[u].size();i++)

        {

            Edge &e = edges[G[u][i]];

            if (e.cap > e.flow && Dis[e.to] > Dis[u]+e.cost)

            {

                Dis[e.to] = Dis[u]+e.cost;

                Pre[e.to] = G[u][i];

                a[e.to] = min(a[u], e.cap-e.flow);

                if (!Vis[e.to])

                {

                    que.push(e.to);

                    Vis[e.to] = 1;

                }

            }

        }

    }

    if (Dis[t] != MINF)

        return true;

    return false;

}

int CostFlow(int &Flow)

{

    int cost = 0;

    while (SPFA())

    {

//        cout << 1 << endl;

//        int Min = INF;

//        for (int i = t;i != s;i = edges[Pre[i]].from)

//            Min = min(Min, edges[Pre[i]].cap-edges[Pre[i]].flow);

//        cout << Min << endl;

        for (int i = t;i != s;i = edges[Pre[i]].from)

        {

            edges[Pre[i]].flow += a[t];

            edges[Pre[i]^1].flow -= a[t];

//            Edge &e = edges[Pre[i]], &ee = edges[Pre[i]^1];

//            cout << e.from << ' ' << e.to << ' ' << e.flow << ' ' << e.cap << endl;

//            cout << ee.from << ' ' << ee.to << ' ' << ee.flow << ' ' << ee.cap << endl;

//            cout << endl;

        }

        cost += a[t]*Dis[t];

        Flow += a[t];

    }

    return cost;

}

void Init()

{

    for (int i = 0;i <= n+m+1;i++)

        G[i].clear();

    edges.clear();

}

int main()

{

//    ios::sync_with_stdio(false);

//    cin.tie(0);

    while (~scanf("%d %d %d", &n, &m, &k) && (n+m+k))

    {

        s = 0, t = n+m+1;

        for (int i = 1;i <= n;i++)

        {

            for (int j = 1;j <= k;j++)

                scanf("%d", &Sh[i][j]);

        }

        memset(SumN, 0, sizeof(SumN));

        for (int i = 1;i <= m;i++)

        {

            for (int j = 1;j <= k;j++)

                scanf("%d", &Wo[i][j]), SumN[j] += Wo[i][j];

        }

        int v;

        //shop = n*k

        //wo = n*k+m*k

        int res = 0, sumflow = 0;

        bool ok = true;

        for (int i = 1;i <= k;i++)

        {

            Init();

            int tmp = 0;

            for (int j = 1;j <= n;j++)

            {

                AddEdge(s, j, Sh[j][i], 0);

                tmp += Sh[j][i];

            }

            if (tmp > SumN[i])

                ok = false;

            for (int j = 1;j <= n;j++)

            {

                for (int z = 1;z <= m;z++)

                {

                    scanf("%d", &v);

                    AddEdge(j, n+z, INF, v);

                }

            }

            for (int j = 1;j <= m;j++)

                AddEdge(n+j, t, Wo[j][i], 0);

            if (ok)

                res += CostFlow(sumflow);

        }

        if (!ok)

            puts("-1");

        else

            printf("%d\n", res);

    }

    return 0;

}

巴特西

POJ-2516-Minimum Cost(网络流, 最小费用最大流)

链接:

题意:

思路:

代码:

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