1227 方格取数 2 费用流

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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;    }
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