POJ3159::SPFA

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 integersN andM not exceeding 30 000 and 150 000 respectively.N is the number of kids in the class and the kids were numbered 1 throughN. snoopy and flymouse were always numbered 1 andN. Then followM lines each holding three integersA, B and c in order, meaning that kidA believed that kidB 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.

Source

POJ Monthly--2006.12.31, Sempr

/*
给定n个糖果,m个条件,和两个糖果a,b之间，w[a]-w[b]的最大差，即w[a]-w[b]<=c
求1到n之间的最大差。

将两个糖果之间的关系看有向边，差看作权值，然后建边，用SPFA即可
参考http://www.cnblogs.com/kuangbin/archive/2012/08/17/2644685.html

2017,11,4
*/

#include 
#include 
#include 
using namespace std;

const int INF = 0x3f3f3f3f;

int n,m;

struct node{
    int to;
    int val;
    int nex;
};
node edge[150005];

int cnt;
int head[30005];
int dis[30005];
bool vis[30005];
int q[30005];

inline void AddEdge(int u,int v,int w)
{
    cnt++;
    edge[cnt].to = v;
    edge[cnt].val = w;
    edge[cnt].nex = head[u];
    head[u] = cnt;
    return;
}

void SPFA()
{
    memset(dis,INF,sizeof(dis));
    
    int top = 1;
    q[top] = 1;
    vis[1] = 1;
    dis[1] = 0;

    while(top != 0)
    {
        int now = q[top];
        top--;
        vis[now] = false;
        for(int i = head[now]; i != -1; i = edge[i].nex)
        {
            int nxt = edge[i].to;
            if(dis[nxt] > dis[now] + edge[i].val)
            {
                dis[nxt] = dis[now] + edge[i].val;
                if(!vis[nxt])
                {
                    vis[nxt] = true;
                    top++;
                    q[top] = nxt;
//                    top++;
                }
            }
        }
    }
}

int main()
{
    memset(head,-1,sizeof(head));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        AddEdge(u,v,w);
    }
    SPFA();
    printf("%d",dis[n]);
}