poj 2516 Minimum Cost--最小费用最大流

/*有N个连锁店，M供货商，供应K中物品，然后还有 某供货商 向 某连锁店 供应 某种物品 的费用输入格式是：N M KN行，每行K个，表示 某连锁店 对于 某物品 的需求M行，每行K个，表示 某供货商 关于 某物品 的储量K个矩阵，每个矩阵N行，M列，表示 某物品 由 某供货商 向 某连锁店 供货的费用输出最小费用，若无法满足，输出-1分别对K中物品进行最小费用最大流就行了在调试本代码的过程中，终于体会到"现在早上5点，那指针现在指向哪儿？"是多么...*/#include<iostream>#include<queue>using namespace std;class solve{public:int n,m,k;int mincost;int** xuqiu;int** gy;int*** fee;int** flow;int** hua;int s,t;int* pre;int* dist;int* vis;void du(){int i,j,h;xuqiu=new int*[n+1];for(i=1;i<=n;++i){xuqiu[i]=new int[k+1];for(j=1;j<=k;++j)cin>>xuqiu[i][j];}gy=new int*[m+1];for(i=1;i<=m;++i){gy[i]=new int[k+1];for(j=1;j<=k;++j)cin>>gy[i][j];}fee=new int**[k+1];for(i=1;i<=k;++i){fee[i]=new int*[n+1];for(j=1;j<=n;++j){fee[i][j]=new int[m+1];for(h=1;h<=m;++h)cin>>fee[i][j][h];}}}solve(int nn,int mm,int kk):n(nn),m(mm),k(kk){int i;s=0;t=n+m+1;mincost=-1;du();chu();if(!manzu()){printf("%d\n",mincost);return;}mincost=0;for(i=1;i<=k;++i){zhuru(i);jiejue(i);}printf("%d\n",mincost);}void jiejue(int w){while(spfa())add();}int spfa()//求费用最小的改进路径      {          dist[s]=0;          for(int i=1;i<n+m+2;i++)              dist[i]=inf();            vis[s]=1;            queue<int>q;          q.push(s);            while(!q.empty())          {              int u=q.front();              for(int v=0;v<=t;++v)              {                  if(flow[u][v]&&dist[v]>dist[u]+hua[u][v])                  {                      dist[v]=dist[u]+hua[u][v];                      pre[v]=u;                        if(!vis[v])                      {                          q.push(v);                          vis[v]=1;                      }                  }              }              q.pop();              vis[u]=0;          }          if(dist[t]==inf())              return 0;          return 1;      }      void add()//压入流      {          int MaxFlow=inf();            int i;                for(i=t;i!=s;i=pre[i])               MaxFlow=min(MaxFlow,flow[pre[i]][i]);                    for(i=t;i!=s;i=pre[i])          {              flow[pre[i]][i]-=MaxFlow;              flow[i][pre[i]]+=MaxFlow;              mincost+=hua[pre[i]][i]*MaxFlow;          }                return;      }  int inf() const{return 0x7FFFFFFF;}//设最大值      int min(int a,int b) {return a<b?a:b;}//取较小的  void zhuru(int w){int i,j;for(i=0;i<n+m+2;++i){memset(flow[i],0,sizeof(int)*(n+m+2));memset(hua[i],0,sizeof(int)*(n+m+2));}for(i=1;i<=m;++i)flow[0][i]=gy[i][w];for(i=1;i<=m;++i)for(j=1;j<=n;++j){hua[i][j+m]=fee[w][j][i];hua[j+m][i]=-hua[i][j+m];flow[i][j+m]=gy[i][w];}for(i=1;i<=n;i++)flow[i+m][t]=xuqiu[i][w];memset(pre,0,sizeof(int)*(n+m+2));memset(vis,0,sizeof(int)*(n+m+2));}int manzu(){int i,j,x,g;for(i=1;i<=k;++i){x=g=0;for(j=1;j<=n;++j)x+=xuqiu[j][i];for(j=1;j<=m;++j)g+=gy[j][i];if(g<x)return 0;}return 1;}void chu(){int i;flow=new int*[n+m+2];hua=new int*[n+m+2];for(i=0;i<n+m+2;++i){flow[i]=new int[n+m+2];hua[i]=new int[n+m+2];}pre=new int[n+m+2];vis=new int[m+n+2];dist=new int[n+m+2];}~solve(){int i,j;delete[] pre;delete[] vis;delete[] dist;for(i=0;i<n+m+2;++i){delete[] flow[i];delete[] hua[i];}delete[] flow;delete[] hua;for(j=1;j<=n;++j)delete[] xuqiu[j];for(j=1;j<=m;++j)delete[] gy[j];delete[] xuqiu;delete[] gy;for(i=1;i<=k;++i){for(j=1;j<=n;++j)delete[] fee[i][j];delete[] fee[i];}delete[] fee;}};int main(){int n,m,k;while(cin>>n>>m>>k,n+m+k)solve poj2516(n,m,k);return 0;}