图论（网络流，二分图最小点权覆盖）：POJ 2125 Destroying The Graph

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 }

 

posted @ 2016-07-07 21:53  TenderRun  阅读(194)  评论(0)  编辑  收藏  举报