maksyuki 发表于 oj 分类,标签: , ,

Destroying The Graph

Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to destroy it. In a move he may take any vertex of the graph and remove either all arcs incoming into this vertex, or all arcs outgoing from this vertex.
Alice assigns two costs to each vertex: Wi+ and Wi-. If Bob removes all arcs incoming into the i-th vertex he pays Wi+ dollars to Alice, and if he removes outgoing arcs he pays Wi- dollars.
Find out what minimal sum Bob needs to remove all arcs from the graph.


Input file describes the graph Alice has drawn. The first line of the input file contains N and M (1 <= N <= 100, 1 <= M <= 5000). The second line contains N integer numbers specifying Wi+. The third line defines Wi- in a similar way. All costs are positive and do not exceed 106 . Each of the following M lines contains two integers describing the corresponding arc of the graph. Graph may contain loops and parallel arcs.


On the first line of the output file print W --- the minimal sum Bob must have to remove all arcs from the graph. On the second line print K --- the number of moves Bob needs to do it. After that print K lines that describe Bob's moves. Each line must first contain the number of the vertex and then '+' or '-' character, separated by one space. Character '+' means that Bob removes all arcs incoming into the specified vertex and '-' that Bob removes all arcs outgoing from the specified vertex.

Sample Input

3 6
1 2 3
4 2 1
1 2
1 1
3 2
1 2
3 1
2 3

Sample Output



1 +

2 -

2 +


Northeastern Europe 2003, Northern Subregion



算法分析:每个点都有两个属性(删除入边的权值和删除出边的权值),则拆点一个处理入边,一个处理出边,使得原图转化为二分图。对于给出的边<u, v>,从u向n+v连一条容量为INF的有向边,从源点向点1~n分别连一条容量为对应点出边点权的有向边,从点n~2n分别向汇点连一条容量为对应点入边点权的有向边。最后跑一个最大流就可得到最小费用值。最后从源点s开始dfs将在残余网络中在同一个连通分量中的点打上标记,所有1~n中没有标记的点就是选择删除出边的,n~2n中被标记的点就是选择删除入边的