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 ",a[i]); 151 else 152 printf("%lld ",w[i]); 153 } 154 return 0; 155 }