bzoj1070: [SCOI2007]修车（费用流）

1070: [SCOI2007]修车

题目：传送门

题解：

　　 一道挺简单的费用流吧...胡乱建模走起

贴个代码...

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 int n,m,st,ed,ans;
 struct node
 {
     int x,y,c,d,next,other;
 }a[];int len,last[];
 void ins(int x,int y,int c,int d)
 {
     int k1,k2;
     k1=++len;
     a[len].x=x;a[len].y=y;a[len].c=c;a[len].d=d;
     a[len].next=last[x];last[x]=len;

     k2=++len;
     a[len].x=y;a[len].y=x;a[len].c=;a[len].d=-d;
     a[len].next=last[y];last[y]=len;

     a[k1].other=k2;
     a[k2].other=k1;
 }
 int head,tail,list[],dis[],pre[],cc[];
 bool v[];
 bool spfa()
 {
     memset(cc,,sizeof(cc));cc[st]=;
     memset(v,false,sizeof(v));v[st]=true;
     memset(dis,0x3F,sizeof(dis));dis[st]=;
     head=;tail=;list[]=st;pre[st]=;
     while(head!=tail)
     {
         int x=list[head];
         for(int k=last[x];k;k=a[k].next)
         {
             int y=a[k].y;
             if( (a[k].c>) && (dis[y]>dis[x]+a[k].d) )
             {
                 cc[y]= min(cc[x],a[k].c);
                 dis[y]=dis[x]+a[k].d;
                 pre[y]=k;
                 if(!v[y])
                 {
                     v[y]=true;
                     list[tail++]=y;
                     if(tail==ed+) tail=;
                 }

             }
         }
         v[x]=false;
         head++; if(head==ed+)head=;
     }
     if(dis[ed]>)return false;
     ans+=dis[ed]*cc[ed];
     int x=ed;
     while(x!=st)
     {
         int k=pre[x];
         a[k].c-=cc[ed];a[a[k].other].c+=cc[ed];
         x=a[k].x;
     }
     return true;
 }
 int map[][];
 int main()
 {
     scanf("%d%d",&m,&n);
     len=;memset(last,,sizeof(last));
     for(int i=;i<=n;i++)
         for(int j=;j<=m;j++)
         {
             scanf("%d",&map[i][j]);
         }
     st=n+m*n+;ed=st+;
     for(int i=;i<=n;i++)
     {
         for(int j=;j<=m;j++)
         {
             for(int k=;k<=n;k++)
             {
                 ins(i,j*n+k,,map[i][j]*(n-k+));
             }
         }
     }
     for(int i=;i<=n;i++)ins(st,i,,);
     for(int i=n+;i<=n+n*m;i++)ins(i,ed,,);
     ans=;
     while(spfa());
     printf("%.2lf\n",double(ans)/n);
     return ;
 }