Candies（差数约分系统-最短路）

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.

n个小孩分糖吃，给出一些约束条件，a,b,x就是b小孩不能比a小孩多大于x的糖果。这样的情况下，问1，n小孩最多差多少糖果。
设糖果数为d，那么就可以转化为 d[b] - d[a] <= x,再转换一下就是d[b] < d[a] + x，对，就是那个最短路的判断条件……然后这题就神奇的转换成了最短路。
转化成yy思想就是，b这个人最多比a多x个糖果，然后他肯定是这些路中取最小的那个。可以画一组样例理解一下
1 2 8
2 3 10
2 4 2
4 3 3

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
ll head[30005],d[30005],n,m;
int st[30005];
bool leap[30005];
struct Node
{
    ll u,v,w,next;
}node[150005];
void add_edge(ll u,ll v,ll w,ll tot)
{
    node[tot].u = u;node[tot].v = v;
    node[tot].w = w;
    node[tot].next = head[u];
    head[u] = tot;
}
void spfa(ll x)
{
    for(ll i = 1;i <= n;i++)d[i] = inf;
    for(ll i = 1;i <= n;i++)leap[i] = false;
    leap[x] = true;d[x] = 0;
    int tot = 0;
    st[tot++] = x;
    while(tot)
    {
        ll u = st[--tot];
        leap[u] = false;
        for(ll i = head[u];i != -1;i = node[i].next)
        {
            ll e = node[i].v;
            if(d[e] > d[u] + node[i].w)
            {
                d[e] = d[u] + node[i].w;
                if(!leap[e])
                {
                    leap[e] = true;st[tot++] = e;
                }
            }
        }
    }

}
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    while(scanf("%d%d",&n,&m) != EOF)
    {
        for(ll i = 1;i <= n;i++)head[i] = -1;
        for(ll i = 1;i <= m;i++)
        {
            ll u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add_edge(u,v,w,i);
        }
        ll temp = 1;
        spfa(temp);
        ll ans = abs(d[1] - d[n]);
        printf("%d\n",ans);
    }
    return 0;
}

然后这题还特妹的卡spfa，手动模拟栈才可以过……