图论（网络流）：SPOJ OPTM - Optimal Marks

OPTM - Optimal Marks

You are given an undirected graph G(V, E). Each vertex has a mark which is an integer from the range [0..231 – 1]. Different vertexes may have the same mark.

For an edge (u, v), we define Cost(u, v) = mark[u] xor mark[v].

Now we know the marks of some certain nodes. You have to determine the marks of other nodes so that the total cost of edges is as small as possible.

Input

The first line of the input data contains integer T (1 ≤ T ≤ 10) - the number of testcases. Then the descriptions of T testcases follow.

First line of each testcase contains 2 integers N and M (0 < N <= 500, 0 <= M <= 3000). N is the number of vertexes and M is the number of edges. Then M lines describing edges follow, each of them contains two integers u, v representing an edge connecting u and v.

Then an integer K, representing the number of nodes whose mark is known. The next K lines contain 2 integers u and p each, meaning that node u has a mark p. It’s guaranteed that nodes won’t duplicate in this part.

Output

For each testcase you should print N lines integer the output. The Kth line contains an integer number representing the mark of node K. If there are several solutions, you have to output the one which minimize the sum of marks. If there are several solutions, just output any of them.

Example

Input:
1
3 2
1 2
2 3
2
1 5
3 100

Output:
5
4
100 

　　COGS上AC了，这里花46分钟买了个教训。
　　SPOJ:
  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <queue>
  5 using namespace std;
  6 const int INF=1000000000;
  7 const int maxn=1010;
  8 const int maxm=30010;
  9 int cnt,fir[maxn],to[maxm],nxt[maxm],cap[maxm];
 10 void addedge(int a,int b,int c){
 11     nxt[++cnt]=fir[a];
 12     fir[a]=cnt;
 13     cap[cnt]=c;
 14     to[cnt]=b;
 15 }
 16 
 17 queue<int>q;
 18 int dis[maxn];
 19 bool BFS(int s,int t){
 20     dis[t]=1;q.push(t);
 21     while(!q.empty()){
 22         int x=q.front();q.pop();
 23         for(int i=fir[x];i;i=nxt[i])
 24             if(!dis[to[i]]){
 25                 dis[to[i]]=dis[x]+1;
 26                 q.push(to[i]);
 27             }
 28     }    
 29     return dis[s];
 30 }
 31 
 32 int fron[maxn];
 33 int gap[maxn],path[maxn];
 34 int ISAP(int s,int t){
 35     if(!BFS(s,t))return 0;
 36     for(int i=s;i<=t;i++)++gap[dis[i]];
 37     for(int i=s;i<=t;i++)fron[i]=fir[i];
 38     int p=s,ret=0,f;
 39     while(dis[s]<=t+10){
 40         if(p==t){
 41             f=INF;
 42             while(p!=s){
 43                 f=min(f,cap[path[p]]);
 44                 p=to[path[p]^1];                
 45             }
 46             ret+=f;p=t;
 47             while(p!=s){
 48                 cap[path[p]]-=f;
 49                 cap[path[p]^1]+=f;
 50                 p=to[path[p]^1];
 51             }
 52         }
 53         int &ii=fron[p];
 54         for(;ii;ii=nxt[ii])
 55             if(cap[ii]&&dis[p]==dis[to[ii]]+1)
 56                 break;
 57         if(ii)
 58             path[p=to[ii]]=ii;
 59         else{
 60             if(--gap[dis[p]]==0)break;
 61             int minn=t+1;
 62             for(int i=fir[p];i;i=nxt[i])
 63                 if(cap[i])minn=min(minn,dis[to[i]]);
 64             ++gap[dis[p]=minn+1];ii=fir[p];
 65             if(p!=s)p=to[path[p]^1];    
 66         }            
 67     }
 68     return ret;
 69 }
 70 
 71 void Init(){
 72     memset(fir,0,sizeof(fir));
 73     memset(dis,0,sizeof(dis));
 74     memset(gap,0,sizeof(gap));
 75     cnt=1;
 76 }
 77 
 78 int n,m,T;
 79 long long a[maxn],w[maxn];
 80 int E[maxm][2],fa[maxn];
 81 int Find(int x){
 82     return fa[x]==x?x:fa[x]=Find(fa[x]);
 83 }
 84 
 85 int vis[maxn];
 86 void DFS(int x,int d){
 87     vis[x]=1;a[x]|=d;
 88     for(int i=fir[x];i;i=nxt[i])
 89         if(cap[i]&&!vis[to[i]])
 90             DFS(to[i],d);
 91 }
 92 
 93 long long Solve(){
 94     int s=0,t=n+1;
 95     long long ret=0;
 96     for(int k=0;k<=30;k++){
 97         Init();
 98         for(int i=1;i<=n;i++)
 99             if(Find(i)==0&&w[i]>=0){
100                 if(w[i]>>k&1){
101                     addedge(s,i,INF);
102                     addedge(i,s,0);
103                 }
104                 else{
105                     addedge(i,t,INF);
106                     addedge(t,i,0);
107                 }
108             }
109         for(int i=1;i<=m;i++)
110             if(Find(E[i][0])==0){
111                 addedge(E[i][0],E[i][1],1);
112                 addedge(E[i][1],E[i][0],1);
113             }
114         ret+=(1ll<<k)*ISAP(s,t);
115         memset(vis,0,sizeof(vis));
116         DFS(s,1ll<<k);
117     }
118     return ret;
119 }
120 
121 int main(){
122     scanf("%d",&T);
123     while(T--){    
124         scanf("%d%d",&n,&m);
125         for(int i=1;i<=m;i++)
126             for(int j=0;j<=1;j++)
127                 scanf("%d",&E[i][j]);
128         
129         int u,v,k;
130         scanf("%d",&k);
131         memset(w,-1,sizeof(w));
132         memset(a,0,sizeof(a));
133         while(k--){
134             scanf("%d",&u);
135             scanf("%lld",&w[u]);
136         }
137         
138         for(int i=1;i<=n;i++)
139             fa[i]=w[i]!=-1?0:i;
140         
141         for(int i=1;i<=m;i++){
142             u=Find(E[i][0]);
143             v=Find(E[i][1]);
144             if(u>v)swap(u,v);
145             if(u!=v)fa[v]=u;
146         }
147         Solve();
148         for(int i=1;i<=n;i++)
149             if(w[i]<0)
150                 printf("%lld\n",a[i]);
151             else 
152                 printf("%lld\n",w[i]);    
153     }
154     return 0;
155 }

 

posted @ 2016-07-04 20:52  TenderRun  阅读(234)  评论(0)  编辑  收藏  举报