hdu 3667 Transportation //拆边费用流

Transportation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 479    Accepted Submission(s): 188

Problem Description

There are N cities, and M directed roads connecting them. Now you want to transport K units of goods from city 1 to city N. There are many robbers on the road, so you must be very careful. The more goods you carry, the more dangerous it is. To be more specific, for each road i, there is a coefficient a _i. If you want to carry x units of goods along this road, you should pay a _i * x ² dollars to hire guards to protect your goods. And what’s worse, for each road i, there is an upper bound C _i, which means that you cannot transport more than C _i units of goods along this road. Please note you can only carry integral unit of goods along each road.
You should find out the minimum cost to transport all the goods safely. 

 

Input

There are several test cases. The first line of each case contains three integers, N, M and K. (1 <= N <= 100, 1 <= M <= 5000, 0 <= K <= 100). Then M lines followed, each contains four integers (u _i, v _i, a _i, C _i), indicating there is a directed road from city u _i to v _i, whose coefficient is a _i and upper bound is C _i. (1 <= u _i, v _i <= N, 0 < a _i <= 100, C _i <= 5)

 

Output

Output one line for each test case, indicating the minimum cost. If it is impossible to transport all the K units of goods, output -1.

 

Sample Input

   
   
   
   

    
    
    
    2 1 2
1 2 1 2
2 1 2
1 2 1 1
2 2 2
1 2 1 2
1 2 2 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
-1
3
   
   
   
   

 

Source

2010 Asia Regional Harbin

 

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#include<cstdio> #include<cstring> const int maxn=150; const int maxm=maxn*maxn;//最大顶点数和边数 const int maxl=999999999; inline int Min(int a,int b) { return a<b?a:b; } inline int Max(int a,int b) { return a>b?a:b; } struct st { int y,d; int ne; int bro; int f; } e[maxm]; int ee,sum; int st[maxn]; void addedge(int x,int y,int d,int f) { //给顶点x和y间添加一条费用d,流量f的边 e[ee].y=y; e[ee].d=d; e[ee].ne=st[x]; e[ee].f=f; st[x]=ee++; e[ee].y=x; e[ee].d=-1*d; e[ee].ne=st[y]; e[ee].f=0; st[y]=ee++; e[ee-2].bro=ee-1; e[ee-1].bro=ee-2; } int d[maxn],p[maxn]; //spfa所用到起点的最短距离(这里距离相当于cost)和路径记录之前的一个节点 int c[maxn];//spfa所用数组:是否在队列中 int que[maxn],head,tail;//spfa专用队列 int spfa(int sx,int ex)//求sx到ex的一次费用增广 { //如果没有增广路就返回maxl 否则返回费用 int i,j,k; for (i=0; i<maxn; i++) d[i]=maxl; memset(c,0,sizeof(c));//初始化都没进 d[sx]=0; que[head=0]=sx; tail=1; c[sx]=1; while (head!=tail) { k=que[head++]; head%=maxn; c[k]=0; for (i=st[k]; i!=-1; i=e[i].ne) if (e[i].f) if (d[k]+e[i].d<d[e[i].y]) { d[e[i].y]=d[k]+e[i].d; p[e[i].y]=i; if (c[e[i].y]==0) { c[e[i].y]=1; if (e[i].d<0) { head=(head-1+maxn)%maxn; que[head]=e[i].y; } else { que[tail++]=e[i].y; tail%=maxn; } } } } if (d[ex]==maxl) return maxl; //如果无法到达终点返回maxl k=maxl; for (i=ex; i!=sx; i=e[e[p[i]].bro].y) k=Min(k,e[p[i]].f); //计算流 sum+=k; for (i=ex; i!=sx; i=e[e[p[i]].bro].y) //增加反向边 { e[p[i]].f-=k; e[e[p[i]].bro].f+=k; } return d[ex]*k; //返回费用为流大小*路径长度(cost累加) } int main() { int n,m,k; while(scanf("%d%d%d",&n,&m,&k)!=EOF) { memset(st,-1,sizeof(st)); ee=0; addedge(0,1,0,k); addedge(n,n+1,0,k); for(int i=1;i<=m;i++) { int a,b,c,d; scanf("%d%d%d%d",&a,&b,&c,&d); if(d==5) { addedge(a,b,c*9,1); d--; } if(d==4) { addedge(a,b,c*7,1); d--; } if(d==3) { addedge(a,b,c*5,1); d--; } if(d==2) { addedge(a,b,c*3,1); d--; } if(d==1) { addedge(a,b,c,1); d--; } } n=n+2; int tot=0,t; sum=0; while ((t=spfa(0,n-1))!=maxl) tot+=t; if(sum==k) printf("%d/n",tot); else printf("-1/n"); } return 0; }