Bzoj1934: [shoi2007] Vote vote in good faith

1934: [shoi2007] Vote goodwill voting time limit: 1 sec memory limit: 64 MB
Submit: 1076 solved: 660
[Submit] [Status] Description there are n children in kindergarten who plan to vote to decide whether to take a nap. For them, this issue is not very important, so they decided to carry forward the spirit of humility. Although everyone has their own opinions, in order to take care of the ideas of their friends, they can also vote for the opposite of their own intentions. We define the number of conflicts in a vote as the total number of conflicts between good friends plus the number of people who conflict with their original intentions. Our question is, how should every child vote to minimize the number of conflicts? The first line of input contains only two integers, N and M, which must be 2 ≤ n ≤ 300,1 ≤ m ≤ n (n-1)/2. N represents the total number of friends, and m represents the logarithm of good friends. There are n integers in the second row of the file. The I integer represents the intention of the I-th child. When it is 1, it indicates that the child agrees to sleep. When it is 0, it indicates that the child is opposed to sleep. The following file contains m rows, each of which has two integers I, j. Indicates that J is a good friend. We guarantee that J will not repeat any two pairs of I. Output only needs to output an integer, that is, the possible minimum number of conflicts. Sample input3 3
1 0 0
1 2
1 3
3 2
Sample output1hint

In the first example, all the children vote in favor to obtain the optimal solution.

Source

Day2

Question: Why is this question so powerful! Worship hzwer: Build source S, sink t
Then S-> xpy with the first vote, xpy with the first vote against T
All traffic is 1
Then add a bidirectional edge for a friend relationship (A, B), and the traffic is still 1
The final cut is the number of conflict
(1) number of conflicts not greater than N:
Obviously, even if there is a friend relationship between all xpy, xpy can reach "agree" or "deny" by changing (or not changing) the original decision ", therefore, the number of conflicts between friends is 0, and the number of conflicts not previously determined is not greater than N.
(2) The edges between the "agree" set and the "deny" set are all friends.
(3) A conflict is a code cut between consent and disagreement:

  1 const inf=maxlongint;  2 type node=record  3      from,go,next,v:longint;  4      end;  5 var  tot,i,j,n,m,maxflow,l,r,s,t,x,y:longint;  6      h,head,q,cur:array[0..1000] of longint;  7      e:array[0..200000] of node;  8      function min(x,y:longint):longint;  9       begin 10       if x<y then exit(x) else exit(y); 11       end; 12 procedure ins(x,y,z:longint); 13  begin 14  inc(tot); 15  e[tot].from:=x;e[tot].go:=y;e[tot].v:=z;e[tot].next:=head[x];head[x]:=tot; 16  end; 17 procedure insert(x,y,z:longint); 18  begin 19  ins(x,y,z);ins(y,x,0); 20  end; 21 function bfs:boolean; 22  var i,x,y:longint; 23  begin 24  fillchar(h,sizeof(h),0); 25  l:=0;r:=1;q[1]:=s;h[s]:=1; 26  while l<r do 27   begin 28   inc(l); 29   x:=q[l]; 30   i:=head[x]; 31   while i<>0 do 32    begin 33    y:=e[i].go; 34    if (e[i].v<>0) and (h[y]=0) then 35     begin 36      h[y]:=h[x]+1; 37      inc(r);q[r]:=y; 38     end; 39    i:=e[i].next; 40    end; 41   end; 42  exit (h[t]<>0); 43  end; 44 function dfs(x,f:longint):longint; 45  var i,y,used,tmp:longint; 46  begin 47  if x=t then exit(f); 48  used:=0; 49  i:=cur[x]; 50  while i<>0 do 51   begin 52   y:=e[i].go; 53   if (h[y]=h[x]+1) and (e[i].v<>0) then 54    begin 55    tmp:=dfs(y,min(e[i].v,f-used)); 56    dec(e[i].v,tmp);if e[i].v<>0 then cur[x]:=i; 57    inc(e[i xor 1].v,tmp); 58    inc(used,tmp); 59    if used=f then exit(f); 60    end; 61   i:=e[i].next; 62   end; 63  if used=0 then h[x]:=-1; 64  exit(used); 65  end; 66 procedure dinic; 67  begin 68  while bfs do 69   begin 70   for i:=s to t do cur[i]:=head[i]; 71   inc(maxflow,dfs(s,inf)); 72   end; 73  end; 74 procedure init; 75  begin 76  tot:=1; 77  readln(n,m); 78  s:=0;t:=n+1; 79  for i:=1 to n do 80   begin 81   read(x); 82   if x=0 then insert(s,i,1) 83   else insert(i,t,1); 84   end; 85  readln; 86  for i:=1 to m do begin readln(x,y);insert(x,y,1);insert(y,x,1);end; 87  end; 88 procedure main; 89  begin 90  maxflow:=0; 91  dinic; 92  writeln(maxflow); 93  end; 94 begin 95  assign(input,‘input.txt‘);assign(output,‘output.txt‘); 96  reset(input);rewrite(output); 97  init; 98  main; 99  close(input);close(output);100 end.       

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