Optimal Marks
Time Limit: 6000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: OPTM
64-bit integer IO format: %lld Java class name: Main
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
Source
解题：amber同学的paper上面有的经典最小割题目
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int INF = ~0U>>2;
4 const int maxn = 510;
5 struct arc{
6 int to,flow,next;
7 arc(int x = 0,int y = 0,int z = -1){
8 to = x;
9 flow = y;
10 next = z;
11 }
12 }e[maxn*maxn];
13 int head[maxn],gap[maxn],d[maxn],S,T,tot;
14 void add(int u,int v,int flow){
15 e[tot] = arc(v,flow,head[u]);
16 head[u] = tot++;
17 e[tot] = arc(u,0,head[v]);
18 head[v] = tot++;
19 }
20 void bfs(){
21 queue<int>q;
22 memset(gap,0,sizeof gap);
23 memset(d,-1,sizeof d);
24 q.push(T);
25 d[T] = 0;
26 while(!q.empty()){
27 int u = q.front();
28 q.pop();
29 ++gap[d[u]];
30 for(int i = head[u]; ~i; i = e[i].next){
31 if(e[i^1].flow && d[e[i].to] == -1){
32 d[e[i].to] = d[u] + 1;
33 q.push(e[i].to);
34 }
35 }
36 }
37 }
38 int sap(int u,int low){
39 if(u == T) return low;
40 int tmp = 0,a,minH = T - 1;
41 for(int i = head[u]; ~i; i = e[i].next){
42 if(e[i].flow){
43 if(d[u] == d[e[i].to] + 1){
44 a = sap(e[i].to,min(low,e[i].flow));
45 if(!a) continue;
46 e[i].flow -= a;
47 e[i^1].flow += a;
48 low -= a;
49 tmp += a;
50 if(!low) break;
51 }
52 minH = min(minH,d[e[i].to]);
53 if(d[S] >= T) return tmp;
54 }
55 }
56 if(!tmp){
57 if(--gap[d[u]] == 0) d[S] = T;
58 ++gap[d[u] = minH + 1];
59 }
60 return tmp;
61 }
62 int maxflow(int ret = 0){
63 bfs();
64 while(d[S] < T) ret += sap(S,INF);
65 return ret;
66 }
67 int n,m,k,mark[maxn],con[maxn];
68 bool mp[maxn][maxn],vis[maxn];
69 void build(int x){
70 S = n + 1;
71 T = S + 1;
72 memset(head,-1,sizeof head);
73 memset(vis,false,sizeof vis);
74 tot = 0;
75 for(int i = 0; i < k; ++i){
76 if((mark[con[i]]>>x)&1) add(S,con[i],INF);
77 else add(con[i],T,INF);
78 }
79 for(int i = 1; i <= n; ++i)
80 for(int j = 1; j <= n; ++j)
81 if(mp[i][j]) add(i,j,1);
82 }
83 void dfs(int u,int x){
84 vis[u] = true;
85 mark[u] |= (1<<x);
86 for(int i = head[u]; ~i; i = e[i].next)
87 if(!vis[e[i].to] && e[i].flow) dfs(e[i].to,x);
88 }
89 int main(){
90 int kase,u,v;
91 scanf("%d",&kase);
92 while(kase--){
93 scanf("%d%d",&n,&m);
94 memset(mark,0,sizeof mark);
95 memset(mp,false,sizeof mp);
96 for(int i = 0; i < m; ++i){
97 scanf("%d%d",&u,&v);
98 mp[u][v] = mp[v][u] = true;
99 }
100 scanf("%d",&k);
101 for(int i = 0; i < k; ++i){
102 scanf("%d%d",&u,&v);
103 mark[u] = v;
104 con[i] = u;
105 }
106 for(int i = 0; i < 32; ++i){
107 build(i);
108 maxflow();
109 dfs(S,i);
110 }
111 for(int i = 1; i <= n; ++i)
112 printf("%d\n",mark[i]);
113 }
114 return 0;
115 }