POJ-3159_Candies

Candies

Time Limit: 1500MS Memory Limit: 131072K

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

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

题意：给n个人派糖果，给出m组数据，每组数据包含A，B，c 三个数，意思是A的糖果数比B少的个数不多于c,即B的糖果数 - A的糖果数<= c 。最后求小孩中的最大差值是多少。

题解：这是一题典型的差分约束题，可以把糖果看成距离，然后就是求最短路的问题了。

由 dis[B]-dis[A]<=w(A,B)。看到这里，我们联想到求最短路时的松弛技术，即if(dis[B]>dis[A]+w(A,B), dis[B]=dis[A]+w(A,B)。

坑点就在于SPFA队列会超时，所以采取堆栈。

#include <iostream>
#include <cstdio>

using namespace std;

const int maxn = 30050;
const int INF = 1e9+7;

int q[maxn],head[maxn],f[maxn],dis[maxn];
int num,n;

struct node
{
    int to,next,w;
}edge[150050];

void add(int v,int u,int w)
{
    edge[num].to = u;
    edge[num].next = head[v];
    edge[num].w = w;
    head[v] = num++;
}

void SPFA()
{
    int i,top;
    for(i=1;i<=n;i++)
    {
        f[i] = 0;
        dis[i] = INF;
    }
    top = 0;
    dis[1] = 0;
    f[1] = 1;
    q[top++] = 1;
    while(top)
    {
        int u,v,w;
        u = q[top-1];
        f[u] = 0;
        top--;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            v = edge[i].to;
            w = edge[i].w;
            if(dis[u]+w<dis[v])
            {
                dis[v] = dis[u] + w;
                if(!f[v])
                {
                    f[v] = 1;
                    q[top++] = v;

                }
            }
        }
    }
    cout<<dis[n]<<endl;
}

int main()
{
    int m,i,a,b,c;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
        head[i] = -1;
    num = 0;
    for(i=0;i<m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);
    }
    SPFA();
    return 0;
}