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
详情请见胡伯涛的论文
经典最小割模型
首先我们分位处理，因为每一位都是独立的，互不影响，然后每个点就变成0或1
再想一下割集的定义，割集把一个图分成了两块，而割集就是这两块之间的连边（好吧，这是我乱编的.....）
xor正好是值不同才会贡献答案，于是建立最小割模型
设一个源s，向确定是1的点连容量为inf的边（因为他不能贡献答案，不能让他成为最小割）
设一个汇t，从确定是0的点向汇连容量为inf的边，理由同上
其他不确定的点就和所有的点（除了s和t）连容量为1的边（双向都要连）
然后跑最大流，也就是最小割
然后我们看这个割集，他把图分成了两部分，最后与s在一起的就是最后为1的（即在残留网络上可以从s走到他），不与s在一起的就是最后为0的，因为是最小割，所以是代价最小的
然后我们要让总和最小，其实我们已经做到了
与s在一起的必须是1，可以画图看一看，不与s在一起的不一定选1，为了和最小，就选0
每一位做一次，最后输出答案就行了
1 const 2 maxn=505; 3 inf=100000; 4 var 5 t,si,ti,n,m,step,time:longint; 6 flag:array[0..maxn]of boolean; 7 tu:array[0..maxn,0..maxn]of boolean; 8 map:array[0..maxn,0..maxn]of longint; 9 a,dis,vh,his,pre,vis:array[0..maxn]of longint; 10 11 procedure sap; 12 var 13 i,j,min,aug:longint; 14 flag:boolean; 15 begin 16 fillchar(dis,sizeof(dis),0); 17 fillchar(vh,sizeof(vh),0); 18 i:=si; 19 vh[0]:=n+2; 20 aug:=inf; 21 while dis[i]<n+2 do 22 begin 23 his[i]:=aug; 24 flag:=false; 25 for j:=1 to ti do 26 if tu[i,j] then 27 if (map[i,j]>0) and (dis[i]=dis[j]+1) then 28 begin 29 if aug>map[i,j] then aug:=map[i,j]; 30 flag:=true; 31 pre[j]:=i; 32 i:=j; 33 if i=ti then 34 begin 35 while i<>si do 36 begin 37 inc(map[i,pre[i]]); 38 dec(map[pre[i],i]); 39 i:=pre[i]; 40 end; 41 aug:=inf; 42 end; 43 break; 44 end; 45 if flag then continue; 46 min:=n+1; 47 for j:=1 to ti do 48 if tu[i,j] then 49 if (map[i,j]>0) and (dis[j]<min) then min:=dis[j]; 50 dec(vh[dis[i]]); 51 if vh[dis[i]]=0 then break; 52 dis[i]:=min+1; 53 inc(vh[dis[i]]); 54 if i<>si then 55 begin 56 i:=pre[i]; 57 aug:=his[i]; 58 end; 59 end; 60 end; 61 62 procedure dfs(x:longint); 63 var 64 i:longint; 65 begin 66 vis[x]:=time; 67 if flag[x]=false then inc(a[x],1<<step); 68 for i:=1 to ti do 69 if vis[i]<>time then 70 if tu[x,i] then 71 if map[x,i]>0 then dfs(i); 72 end; 73 74 procedure main; 75 var 76 j,k,m,x,y:longint; 77 begin 78 fillchar(flag,sizeof(flag),false); 79 fillchar(tu,sizeof(tu),false); 80 fillchar(a,sizeof(a),0); 81 read(n,m); 82 si:=0; 83 ti:=n+1; 84 for j:=1 to m do 85 begin 86 read(x,y); 87 tu[y,x]:=true; 88 tu[x,y]:=true; 89 end; 90 for j:=1 to n do 91 begin 92 tu[si,j]:=true; 93 tu[j,ti]:=true; 94 end; 95 read(m); 96 for j:=1 to m do 97 begin 98 read(x,y); 99 flag[x]:=true; 100 a[x]:=y; 101 end; 102 for step:=0 to 30 do 103 begin 104 fillchar(map,sizeof(map),0); 105 for j:=1 to n do 106 if flag[j] then 107 if a[j] and (1<<step)=0 then 108 begin 109 map[j,ti]:=inf; 110 for k:=1 to n do 111 if flag[k]=false then map[k,j]:=1; 112 end 113 else 114 begin 115 map[si,j]:=inf; 116 for k:=1 to n do 117 if flag[k]=false then map[j,k]:=1; 118 end 119 else 120 for k:=j+1 to n do 121 if flag[k]=false then 122 begin 123 map[j,k]:=1; 124 map[k,j]:=1; 125 end; 126 sap; 127 inc(time); 128 dfs(0); 129 end; 130 for j:=1 to n do 131 writeln(a[j]); 132 end; 133 134 begin 135 read(t); 136 while t>0 do 137 begin 138 main; 139 dec(t); 140 end; 141 end.