[原]1227 方格取数 2 费用流

楚东方 17/10/11 00:10:24

题目链接

要点,每个点之间可以连两条边,这样可以解决很多问题,该题中的数据被取出,则可以连一条有权重的边和若干条无权重的边.

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;


    const int MAXN = 10000;
    const int MAXM = 1000000;
    const int INF = 0x3f3f3f3f;
    struct Edge
    {
        int to,next,cap,flow,cost;
    } edge[MAXM];
    int head[MAXN],tol;
    int pre[MAXN],dis[MAXN];
    bool vis[MAXN];
    int N;//节点总个数,节点编号从0~N-1
    void init(int n)
    {
        N = n;
        tol = 0;
        memset(head,-1,sizeof(head));
    }
    void addedge(int u,int v,int cap,int cost)
    {
        edge[tol].to = v;
        edge[tol].cap = cap;
        edge[tol].cost = cost;
        edge[tol].flow = 0;
        edge[tol].next = head[u];
        head[u] = tol++;
        edge[tol].to = u;
        edge[tol].cap = 0;
        edge[tol].cost = -cost;
        edge[tol].flow = 0;
        edge[tol].next = head[v];
        head[v] = tol++;
    }
    bool spfa(int s,int t)
    {

        queue<int>q;
        for(int i = 0; i < N; i++)
        {
            dis[i] = INF;
            vis[i] = false;
            pre[i] = -1;
        }
        dis[s] = 0;
        vis[s] = true;
        q.push(s);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            vis[u] = false;
            for(int i = head[u]; i != -1; i = edge[i].next)
            {
                int v = edge[i].to;
                if(edge[i].cap > edge[i].flow &&
                        dis[v] > dis[u] + edge[i].cost )
                {
                    dis[v] = dis[u] + edge[i].cost;
                    pre[v] = i;
                    if(!vis[v])
                    {
                        vis[v] = true;
                        q.push(v);
                    }
                }
            }
        }
        if(pre[t] == -1)return false;
        else return true;
    }
    //返回的是最大流,cost存的是最小费用
    int minCostMaxflow(int s,int t,int &cost)
    {
        int flow = 0;
        cost = 0;
        while(spfa(s,t))
        {
            int Min = INF;
            for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
            {
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            }
            for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
                cost += edge[i].cost * Min;
            }
            flow += Min;

        }
        return flow;
    }


    int n,k;
    int mm[100][100];



    int main()
    {
    //    freopen("data.txt","r",stdin);
        ios_base::sync_with_stdio(false);
        cin >> n>>k;
        int s = 0;
        int t  = n*n*2+1;
        init(n*n*2+10);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin >> mm[i][j];
            }
        }


        addedge(s,1,k,0);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                int num =(i-1)*n+j;
                addedge( num,num + n*n, 1  ,-mm[i][j] );
                addedge( num,num + n*n, INF,0 );
            }
        }

        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                int fr =(i-1)*n+j + n*n;
                int to =(i)*n+j;
                if(i!=n)
                    addedge(fr,to,INF,0 );

                fr =(i-1)*n+j + n*n;
                to =(i-1)*n+j+1;
                if(j!=n)
                    addedge(fr,to,INF,0 );
            }
        }

        addedge( 2*n*n,t,INF,0);

        int ans  =0;
        minCostMaxflow(s,t,ans);
        cout << -ans << endl;

        return 0;
    }



















作者:chudongfang2015 发表于2017/10/11 0:10:24 原文链接
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