King - UVa 515 差分约束系统

King

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 

 of
a sequence 

. The king thought a minute and then decided, i.e. he set for the sum 

 of
each subsequence Si an integer constraint ki (i.e. 

 or 

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 file 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 

 is
length of the sequence S and 

 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 file contains the lines corresponding to the blocks in the input file. A line contains textsuccessful conspiracy when such a sequence
does not exist. Otherwise it contains text lamentable kingdom. There is no line in the output file corresponding to the last ``null'' block of the input file.

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

题意：需要你自己去构造一个序列，满足他给定的要求，即从s到s+n的数的和大于或小于给定的k。问是否存在这样的序列。

思路：首先将序列变成X0，X1，X2...Xn的形式，表示前i项的和，那么每个约束条件我们都可以变成Xi-Xj<=k的形式，然后构图，使Xj指向Xi的有向边的权值为k，另外再找到一个点n+1，使得其到所有点的有向边的权值为0。如果这个图没有负环存在的话，那么就是有解的。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int n,m,d[110],vis[110],p[110],INF=1e9;
char s[10];
queue<int> qu;
struct node
{
int v,w,next;
}edge[510];
int h[110],tot;
void AddEdge(int u,int v,int w)
{
edge[++tot].v=v;
edge[tot].w=w;
edge[tot].next=h[u];
h[u]=tot;
}
void spfa(int s)
{
int i,j,k,u,v,w,f;
memset(vis,0,sizeof(vis));
memset(p,0,sizeof(p));
for(i=0;i<=n;i++)
d[i]=INF;
d[s]=0;
while(!qu.empty())
qu.pop();
qu.push(s);vis[s]=1;
while(!qu.empty())
{
u=qu.front();
qu.pop();
for(f=h[u];f;f=edge[f].next)
{
v=edge[f].v;
w=edge[f].w;
if(d[u]+w<d[v])
{
d[v]=d[u]+w;
if(vis[v]==0)
{
qu.push(v);
vis[v]=1;
if(++p[v]>n)
{
printf("successful conspiracy\n");
return;
}
}
}
}
vis[u]=0;
}
printf("lamentable kingdom\n");
}
int main()
{
int i,j,k,a,b,u,v;
while(~scanf("%d",&n) && n>0)
{
scanf("%d",&m);
memset(h,0,sizeof(h));tot=0;
n++;
for(i=1;i<=m;i++)
{
scanf("%d%d%s%d",&a,&b,s,&k);
if(s[0]=='g')
AddEdge(a+b,a-1,-k-1);
else
AddEdge(a-1,a+b,k-1);
}
for(i=0;i<=n;i++)
AddEdge(n,i,0);
spfa(n);
}
}