poj 1364

本文介绍了一种利用SPFA算法解决差分约束系统的方法,通过将约束条件转化为差分形式,实现对特定序列的存在性判断。

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Description

Once, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: “If my child was a son and if only he was a sound king.” After nine months her child was born, and indeed, she gave birth to a nice son.
Unfortunately, as it used to happen in royal families, the son was a little retarded. After many years of study he was able just to add integer numbers and to compare whether the result is greater or less than a given integer number. In addition, the numbers had to be written in a sequence and he was able to sum just continuous subsequences of the sequence.

The old king was very unhappy of his son. But he was ready to make everything to enable his son to govern the kingdom after his death. With regards to his son’s skills he decided that every problem the king had to decide about had to be presented in a form of a finite sequence of integer numbers and the decision about it would be done by stating an integer constraint (i.e. an upper or lower limit) for the sum of that sequence. In this way there was at least some hope that his son would be able to make some decisions.

After the old king died, the young king began to reign. But very soon, a lot of people became very unsatisfied with his decisions and decided to dethrone him. They tried to do it by proving that his decisions were wrong.

Therefore some conspirators presented to the young king a set of problems that he had to decide about. The set of problems was in the form of subsequences Si = {aSi, aSi+1, …, aSi+ni} of a sequence S = {a1, a2, …, an}. The king thought a minute and then decided, i.e. he set for the sum aSi + aSi+1 + … + aSi+ni of each subsequence Si an integer constraint ki (i.e. aSi + aSi+1 + … + aSi+ni < ki or aSi + aSi+1 + … + aSi+ni > ki resp.) and declared these constraints as his decisions.

After a while he realized that some of his decisions were wrong. He could not revoke the declared constraints but trying to save himself he decided to fake the sequence that he was given. He ordered to his advisors to find such a sequence S that would satisfy the constraints he set. Help the advisors of the king and write a program that decides whether such a sequence exists or not.
Input

The input consists of blocks of lines. Each block except the last corresponds to one set of problems and king’s decisions about them. In the first line of the block there are integers n, and m where 0 < n <= 100 is length of the sequence S and 0 < m <= 100 is the number of subsequences Si. Next m lines contain particular decisions coded in the form of quadruples si, ni, oi, ki, where oi represents operator > (coded as gt) or operator < (coded as lt) respectively. The symbols si, ni and ki have the meaning described above. The last block consists of just one line containing 0.
Output

The output contains the lines corresponding to the blocks in the input. A line contains text successful conspiracy when such a sequence does not exist. Otherwise it contains text lamentable kingdom. There is no line in the output corresponding to the last “null” block of the input.
Sample Input

4 2
1 2 gt 0
2 2 lt 2
1 2
1 0 gt 0
1 0 lt 0
0
Sample Output

lamentable kingdom
successful conspiracy
Source

Central Europe 1997

分析:
题意:相当的晦涩,大概就是对于一个序列数字序列S。给出(si, ni, 0, ki)。
如果是(si,ni,gt,ki),意思就是存在约束条件S[si]+S[si+1]+…S[si+ni] > ki,
如果是(si,ni,lt,ki),意思就是存在约束条件S[si]+S[si+1]+…S[si+ni] < ki
判断所给的约束条件有无解,即是否存在这么一个序列S,有解就输出lamentable kingdom,无解就输出successful conspiracy。

思路:spfa+差分约束。
将这些约束条件转化为差分约束,不妨设T[x] = S[1]+S[2]+….S[x],那么上面式子就可以转化为:
1. T[si+ni] - T[si-1] > ki 2. T[si+ni] - T[si-1] < ki
又差分约束系统的求解只能针对>=或<=,无法求解>的系统,还好ki是整数,可以很容易将<转化为<=
1. T[si-1] - T[si+ni] <= -ki - 1
2. T[si+ni] - T[si-1] <= ki - 1
剩下的就是普通的差分约束系统的求解了,一种无需建立超级源点的SPFA解法:在SPFA开始时将所有结点都放进队列,这样表示一开始和所有点都相连了,初始化dis数组为全0,相当于超级源点的边权值。

代码:

const
  MaxE=102;
  MaxV=500001;

type
  rec=record
   x,y,w,next:longint;
  end;
var
  n,m,c,qx,i,x,y,w,o,q,p:longint;
  g:array [-1..Maxv] of rec;
  ls:array [-1..Maxe] of longint;
  v,d,list,sum:array [-1..maxe] of longint;
  ch:string;
  boo:boolean;

procedure spfa(first:longint);
var
  head,tail,t,i,qe:longint;
begin
  tail:=1;
  list[1]:=first;
  for i:=1 to n do
  d[i]:=maxlongint;
  d[first]:=0;
  v[first]:=1;
  while tail<>0 do
    begin
      t:=ls[list[tail]];
      qe:=list[tail];
      tail:=tail-1;
      while t>0 do
        with g[t] do
          begin
            if d[x]+w<d[y] then
              begin
                d[y]:=d[x]+w;
                if v[y]=0 then
                  begin
                    v[y]:=1;
                    tail:=tail+1;
                    list[tail]:=y;
                    inc(sum[y]);
                    if (sum[y]>=n) then
                     begin
                      boo:=false;
                      exit;
                     end;
                  end;
              end;
            t:=next;
          end;
      v[qe]:=0;
    end;
end;

procedure add(x,y,w:longint);
 begin
  inc(o);
  g[o].x:=x;
  g[o].y:=y;
  g[o].w:=w;
  g[o].next:=ls[x];
  ls[x]:=o;
 end;

begin
 read(n);
while n>0 do
 begin
  readln(m);
  for i:=1 to m do
   begin
    x:=0; y:=0; w:=0;
    readln(ch);
    p:=pos(' ',ch);
    val(copy(ch,1,p-1),x);
    delete(ch,1,p);
     p:=pos(' ',ch);
    val(copy(ch,1,p-1),y);
    delete(ch,1,p);
    p:=pos(' ',ch);
    val(copy(ch,p+1,length(ch)),w);
    if ch[1]='g' then add(x-1,x+y,-w-1)
                 else add(x+y,x-1,w-1);
   end;
  for i:=0 to n do
  add(-1,i,0);
  boo:=true;
  spfa(-1);
  fillchar(g,sizeof(g),0);
  fillchar(ls,sizeof(ls),0);
  o:=0;
  fillchar(v,sizeof(v),0);
  fillchar(d,sizeof(d),$7f);
  fillchar(sum,sizeof(sum),0);
  fillchar(list,sizeof(list),0);
  if boo then writeln('lamentable kingdom')
         else writeln('successful conspiracy');
  read(n);
 end;
end.

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