poj3159 Candies

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.

差分约束系统模板题。
对于每个条件y<=x+k网上许多说法是把x，y看成点，但是我认为更准确的说法是把x，y看成单源最短路的距离。也就是dis[y]<=dis[x]+k，那么连一条y到x长度为k的边，这样才更准确。谁是源呢？这道题里，显然1是源。
这样建好图以后跑单源最短路即可。
注意这题卡spfa，最后堆优化dij跑了将近0.6s。

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
struct node
{
    int num,dis;
    bool operator < (const node &a) const
    {
        return dis>a.dis;
    }
}n1,n2;
priority_queue<node> q;
int dis[30010],fir[30010],ne[150010],to[150010],len[150010];
bool vis[30010];
int main()
{
    int i,j,k,m,n,p,x,y,z,ans;
    scanf("%d%d",&n,&m);
    for (i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        ne[i]=fir[x];
        fir[x]=i;
        len[i]=z;
        to[i]=y;
    }
    memset(dis,0x3f,sizeof(dis));
    dis[1]=0;
    q.push((node){1,0});
    while (!q.empty())
    {
        n1=q.top();
        q.pop();
        x=n1.num;
        if (vis[x]) continue;
        vis[x]=1;
        for (i=fir[x];i;i=ne[i])
          if (dis[x]+len[i]<dis[to[i]])
          {
            dis[to[i]]=dis[x]+len[i];
            q.push((node){to[i],dis[to[i]]});
          }
    }
    printf("%d\n",dis[n]); 
}