Candies POJ

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 cin 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

32-bit signed integer type is capable of doing all arithmetic.

本题让让求的是最长路径

而由于需要将每个条件都满足

可以把条件转换为松弛条件

从而转化为最短路的问题

本题我用了spfa但是使用队列的话会超时

之前听说有专门卡是spfa的数据可能这题就是了

但有一个小技巧

可以将spfa的队列操作改为stack

这样恶心的数据可能就被弱化了 当然要卡应该还是会被卡掉

//²î·ÖÔ¼Êø
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
int n,m;
int dis[30005];
int first[150005];
int nex[150005];
int vis[30005];
struct node
{
    int to;
    int w;
}a[150005];
stack <int> q;
int spfa()
{
    int temp;
    dis[1]=0;
    q.push(1);
    vis[1]=1;
    while(q.size())
    {
        temp=q.top();
        q.pop();
        vis[temp]=0;
        for(int i=first[temp];i!=-1;i=nex[i])
        {
            if(dis[a[i].to]>dis[temp]+a[i].w)
            {
                dis[a[i].to]=dis[temp]+a[i].w;
                if(vis[a[i].to]==0)
                {
                    vis[a[i].to]=1;
                    q.push(a[i].to);
                }
            }
        }
    }
    return dis[n];
}
int main()
{
    memset(vis,0,sizeof(vis));
    memset(first,-1,sizeof(first));
    memset(nex,-1,sizeof(nex));
    memset(dis,0x3f,sizeof(dis));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int temp1,temp2,temp3;
        scanf("%d%d%d",&temp1,&temp2,&temp3);
        nex[i]=first[temp1];
        first[temp1]=i;
        a[i].w=temp3;
        a[i].to=temp2;
    }
    printf("%d
", spfa());
}