SPOJ OPTM Optimal Marks

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

Guo HuaYang

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
",mark[i]);
113     }
114     return 0;
115 }

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