K - Candies POJ - 3159

K - Candies

POJ - 3159

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

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

 

题意：分配糖果，A,B,c表示A最多比B少多少糖果，求最后n比1最多能多多少糖果。

题解：dis[B]表示B最多分配糖果，dis[A]表示A最多分配糖果，那么dis[B] - dis[A] <= c，在最短路中的松弛操作为:

         if(dis[B] > dis[A] + c)       dis[B] = dis[A] + c ,这题我用队列超时，看网上说用栈就不超时，我也不知道为什么栈比队列快。

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 151000;
int dis[maxn],head[maxn],inq[maxn],n,m,tot;

struct node
{
    int v;
    int w;
    int next;
}edg[maxn];

void addnode(int u,int v,int w)
{
    edg[tot].v = v;
    edg[tot].w = w;
    edg[tot].next = head[u];
    head[u] = tot++;
}

void SPFA()
{
    memset(dis,inf,sizeof(dis));
    memset(inq,0,sizeof(inq));

    stackQ;
    Q.push(1);
    inq[1] = 1;
    dis[1] = 0;

    while(!Q.empty())
    {
        int u = Q.top();
        Q.pop();
        inq[u] = 0;
        for(int i = head[u];i != -1;i = edg[i].next)
        {
            int v = edg[i].v,w = edg[i].w;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if(!inq[v])
                {
                    Q.push(v);
                    inq[v] = 1;
                }
            }
        }
    }
}

int main()
{
    scanf("%d %d",&n,&m);

    memset(head,-1,sizeof(head));
    tot = 0;

    for(int i = 1;i <= m;i++)
    {
        int u,v,w;
        scanf("%d %d %d",&u,&v,&w);
        addnode(u,v,w);
    }

    SPFA();

    printf("%d\n",dis[n]);

    return 0;
}