| Time Limit: 1500MS | Memory Limit: 131072K | |
| Total Submissions: 12484 | Accepted: 3226 |
Description
During the kindergarten days, flymouse was the monitor of his class.
Occasionally the head-teacher brought the kids of flymouse’s class a large bag
of candies and had flymouse distribute them. All the kids loved candies very
much and often compared the numbers of candies they got with others. A kid A
could had the idea that though it might be the case that another kid B was
better than him in some aspect and therefore had a reason for deserving more
candies than he did, he should never get a certain number of candies fewer than
B did no matter how many candies he actually got, otherwise he would feel
dissatisfied and go to the head-teacher to complain about flymouse’s biased
distribution.
snoopy shared class with flymouse at that time. flymouse always compared the
number of his candies with that of snoopy’s. He wanted to make the difference
between the numbers as large as possible while keeping every kid satisfied. Now
he had just got another bag of candies from the head-teacher, what was the
largest difference he could make out of it?
Input
The input contains a single test cases. The test cases starts with a line
with two integers N and M not exceeding 30 000 and 150 000
respectively. N is the number of kids in the class and the kids were
numbered 1 through N. snoopy and flymouse were always numbered 1 and
N. Then follow M lines each holding three integers A,
B and c in order, meaning that kid A believed that kid
B should never get over c candies more than he did.
Output
Output one line with only the largest difference desired. The difference is
guaranteed to be finite.
Sample Input
2 2 1 2 5 2 1 4
Sample Output
5
Hint
arithmetic.
Source
Monthly--2006.12.31, Sempr
const int INF=1<<30-1;
const int E=150001;
const int V=30001;
struct EDGE
{
int link,val;
int next;
}edge[E];
int head[V],dist[V],e;
bool vis[V];
inline void addedge(int a,int b,int c)
{
edge[e].link=b;
edge[e].val=c;
edge[e].next=head[a];
head[a]=e++;
}
int relax(int u,int v,int c)
{
if(dist[v]>dist[u]+c)
{
dist[v]=dist[u]+c;
return 1;
}
return 0;
}
int SPFA(int src,int n)
{
for(int i=1;i<=n;i++)
{
vis[i]=0;
dist[i]=INF;
}
dist[src]=0;
vis[src]=true;
int Q[E],top=1;
Q[0]=src;
while(top)
{
int u,v;
u=Q[--top];
vis[u]=false;//还可以放进来
for(int i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].link;
if(relax(u,v,edge[i].val)==1&&!vis[v] )
{
Q[top++]=v;
vis[v]=true;
}
}
}
return dist[n];
}