題目大意：給出N個孩子，M個要求，問滿足所有孩子的要求時，第一個孩子的糖果和第N個孩子的糖果差的最大值是多少

解題思路：差分約束系統的裸題，B孩子的糖果數量- A孩子的糖果數量 <= C，即d[B] - d[A] <= C，所以可以構出一條A指向B的邊，權值為C
跑一遍SPFA，注意要用棧代替佇列，不然會TLE

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define N 30010
#define M 150010
 

#define INF 0x3f3f3f3f

struct Edge{
    int to, dist, next;
}E[M];

int head[N], d[N], n, m, tot;
int stack[N];
bool vis[N];

void AddEdge(int from, int to, int dist) {
    E[tot].to = to;
    E[tot].dist = dist;
    E[tot].next = head[from];
    head[from] = tot++;
}

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

    int u, v, d;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &u, &v, &d);
        AddEdge(u, v, d);
    }
}

void SPFA() {
    for (int i = 2; i <= n; i++) 
        d[i] = INF;
    d[1] = 0;
    int top = 0;
    stack[++top] = 1;
    vis[1] = true
 
;

    while (top) {
        int u = stack[top--];

        for (int i = head[u]; i != -1; i = E[i].next) {
            int v = E[i].to;
            if (d[v] > d[u] + E[i].dist) {
                d[v] = d[u] + E[i].dist;
                if (!vis[v]) {
                    vis[v] = true;
                    stack[++top] = v;
                }
            }
        }
        vis[u] = false;
    }
    printf("%d\n", d[n]);
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        init();
        SPFA();
    }
    return 0;
}