蚯蚓的游戏问题

#include<bits/stdc++.h>
#define N 2700
#define INF LLONG_MAX/2

using namespace std;


struct MCMF
{
    struct Edge
    {
        long long  from,to,flow,cost;
    };

    vector<Edge> edges;
    vector<long long > G[N];
    long long  inq[N];//是否在队列中
    long long  d[N];//距离
    long long  p[N];//上一条弧
    long long  a[N];//可改进量

    void init()//初始化
    {
        for(long long  i=0; i<N; i++)G[i].clear();
        edges.clear();
    }

    void addedge(long long  from,long long  to,long long  flow,long long  cost)//加边
    {
        edges.push_back((Edge)
        {
            from,to,flow,cost
        });
        edges.push_back((Edge)
        {
            to,from,0,-cost
        });
        long long  m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool SPFA(long long  s,long long  t,long long  &flow,long long  &cost)//寻找最小费用的增广路，使用引用同时修改原flow,cost
    {
        for(long long  i=0; i<N; i++)d[i]=INF,inq[i]=0;
        d[s]=0;
        inq[s]=1;
        p[s]=0;
        a[s]=INF;
        queue<long long > Q;
        Q.push(s);
        while(!Q.empty())
        {
            long long  u=Q.front();
            Q.pop();
            inq[u]=0;
            for(long long  i=0; i<G[u].size(); i++)
            {
                Edge& e=edges[G[u][i]];
                if(e.flow>0 && d[e.to]>d[u]+e.cost)//满足可增广且可变短
                {
                    d[e.to]=d[u]+e.cost;
                    p[e.to]=G[u][i];
                    a[e.to]=min(a[u],e.flow);
                    if(!inq[e.to])
                    {
                        inq[e.to]=1;
                        Q.push(e.to);
                    }
                }
            }
        }
        if(d[t]==INF) return false;//汇点不可达则退出
        flow+=a[t];
        cost+=d[t]*a[t];
        long long  u=t;
        while(u!=s)//更新正向边和反向边
        {
            edges[p[u]].flow-=a[t];
            edges[p[u]^1].flow+=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }


    void MincotMaxflow(long long  s,long long  t,long long  &flow,long long  &cost)
    {
        while(SPFA(s,t,flow,cost));//{cost=0;flow=long long _MAX;}
    }
};

MCMF mcmf;
int main()
{
    int n,m,k;
    int a[40][80];
    int x[40][80],y[40][80];

    scanf("%d %d %d",&n,&m,&k);

    for(int i=1; i<=n; i++)
        for(int j=1; j<=m+i-1; j++)scanf("%d",&a[i][j]);

    int s2=((m+m+n-1)*n/2)*2+1;
    int s1=s2+1;
    int t=s1+1;

    mcmf.init();
    mcmf.addedge(s1,s2,k,0);

    for(int i=1; i<=m; i++)mcmf.addedge(s2,2*i-1,1,0);

    int tot=1;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m+i-1; j++)
        {
            x[i][j]=2*tot-1,y[i][j]=2*tot;
            tot++;
        }
    }


    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m+i-1; j++)
            mcmf.addedge(x[i][j],y[i][j],1,-a[i][j]);
    }

    for(int i=1; i<n; i++)
    {
        for(int j=1; j<=m+i-1; j++)
        {
            mcmf.addedge(y[i][j],x[i+1][j],1,0);
            mcmf.addedge(y[i][j],x[i+1][j+1],1,0);
        }
    }

    for(int i=1; i<=n+m+1; i++)
    {
        mcmf.addedge(y[n][i],t,1,0);
    }

    long long flow=0,cost=0;
    mcmf.MincotMaxflow(s1,t,flow,cost);
    printf("%lld",-cost);
    return 0;
}