Destroying The Graph
Description
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
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.
Output
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
5 3 1 + 2 - 2 +
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <queue>
5 using namespace std;
6 const int maxn=1010;
7 const int maxm=100010;
8 const int INF=1000000000;
9 int cnt,tot,fir[maxn],fron[maxn],dis[maxn];
10 int to[maxm],nxt[maxm],gap[maxn],path[maxn];
11 int cap[maxm];queue<int>q;
12
13 struct Max_Flow{
14 void Init(int tot_=0){
15 tot=tot_;cnt=1;
16 memset(fir,0,sizeof(fir));
17 memset(dis,0,sizeof(dis));
18 memset(gap,0,sizeof(gap));
19 }
20
21 void add(int a,int b,int c){
22 nxt[++cnt]=fir[a];
23 fir[a]=cnt;
24 cap[cnt]=c;
25 to[cnt]=b;
26 }
27
28 void addedge(int a,int b,int c){
29 add(a,b,c);
30 add(b,a,0);
31 }
32
33 bool BFS(int s,int t){
34 dis[t]=1;q.push(t);
35 while(!q.empty()){
36 int x=q.front();q.pop();
37 for(int i=fir[x];i;i=nxt[i])
38 if(!dis[to[i]]){
39 dis[to[i]]=dis[x]+1;
40 q.push(to[i]);
41 }
42 }
43 return dis[s];
44 }
45
46 int Aug(int s,int t,int &p){
47 int f=INF;
48 while(p!=s){
49 f=min(f,cap[path[p]]);
50 p=to[path[p]^1];
51 }p=t;
52 while(p!=s){
53 cap[path[p]]-=f;
54 cap[path[p]^1]+=f;
55 p=to[path[p]^1];
56 }
57 return f;
58 }
59
60 int ISAP(int s,int t){
61 if(!BFS(s,t))return 0;
62 for(int i=s;i<=t;i++)fron[i]=fir[i];
63 for(int i=s;i<=t;i++)gap[dis[i]]+=1;
64 int p=s,ret=0;
65 while(dis[s]<=tot){
66 if(p==t)ret+=Aug(s,t,p);
67
68 for(int &i=fron[p];i;i=nxt[i])
69 if(cap[i]&&dis[p]==dis[to[i]]+1){
70 path[p=to[i]]=i;
71 break;
72 }
73
74 if(!fron[p]){
75 if(--gap[dis[p]]==0)
76 break;
77 int Min=tot;
78 for(int i=fir[p];i;i=nxt[i])
79 if(cap[i])Min=min(Min,dis[to[i]]);
80 gap[dis[p]=Min+1]+=1;fron[p]=fir[p];
81 if(p!=s)p=to[path[p]^1];
82 }
83 }
84 return ret;
85 }
86 }isap;
87
88 int n,m,top;
89 int tag[maxn],st[maxn];
90 void DFS(int x){
91 tag[x]=1;
92 for(int i=fir[x];i;i=nxt[i])
93 if(cap[i]&&!tag[to[i]])DFS(to[i]);
94 }
95
96 int main(){
97 scanf("%d%d",&n,&m);
98 int s=0,t=2*n+1;
99 isap.Init(t+1);
100 for(int i=1,v;i<=n;i++){
101 scanf("%d",&v);
102 isap.addedge(s,i,v);
103 }
104 for(int i=1,v;i<=n;i++){
105 scanf("%d",&v);
106 isap.addedge(i+n,t,v);
107 }
108 for(int i=1,a,b;i<=m;i++){
109 scanf("%d%d",&a,&b);
110 isap.addedge(b,a+n,INF);
111 }
112
113 printf("%d\n",isap.ISAP(s,t));
114 DFS(0);
115 for(int i=1;i<=n;i++){
116 if(!tag[i])
117 st[++top]=i;
118 if(tag[i+n])
119 st[++top]=i+n;
120 }
121 printf("%d\n",top);
122 for(int i=1;i<=top;i++){
123 if(st[i]<=n)
124 printf("%d +\n",st[i]);
125 else
126 printf("%d -\n",st[i]-n);
127 }
128 return 0;
129 }