poj 3159 Candies 差分约束

        设a[i] 为第i个小孩得到的糖果数，d[i] 为第i个小孩相对于第1个小孩的糖果数，即有：

        d[i] = a[i] – a[1]， d[1] = 0.

       题目输入的A B c 有： a[B] – a[A] <= c  即有 (a[B] – a[1]) – (a[A] – a[1]) <= c ， d[B] – d[A] <= c

       题目是求最大值，就是求以1为源点的最短路。

       注意：用spfa+队列会tlm, 要改用stack才过, 700ms。试用了Dijkstra+priority_queue，1400+ms擦过去了，不知是不是写得不够好.

 

#include <iostream>
#include <queue>
#include <stack>
using namespace std;

const int MAX = 30005;
const int INF = 1000000000;
const int N = 150005;

struct Node
{
	int v;
	int cost;
	int next;
};

Node node[N];
int d[MAX];
int adj[MAX];
bool in_q[MAX];
int cnt[MAX];
int size;
int n, m;

void add_edge(int u, int v, int cost)
{
	node[size].v = v;
	node[size].cost = cost;
	node[size].next = adj[u];
	adj[u] = size++;
}

struct cmp
{
	bool operator() (const int &a, const int &b)
	{
		return d[a] > d[b];
	}
};

priority_queue<int, vector<int>, cmp> Q;

void Dijkstra()
{
	for (int i = 0; i <= n; i++)
		d[i] = INF;

	
	Q.push(1);
	d[1] = 0;

	int u, v, w;
	while (!Q.empty())
	{
		u = Q.top();
		Q.pop();

		for (int i = adj[u]; i != -1; i = node[i].next)
		{
			v = node[i].v;
			w = node[i].cost;

			if (d[v] > d[u] + w)
			{
				d[v] = d[u] + w;
				Q.push(v);
			}
		}
	}
}

void spfa()
{
	memset(in_q, false, sizeof(in_q));

	for (int i = 1; i <= n; i++)
		d[i] = INF;

	stack<int> S;
	d[1] = 0;
	S.push(1);
	in_q[1] = true;
	int u, v, w;

	while (!S.empty())
	{
		u = S.top();
		S.pop();
		in_q[u] = false;

		for (int i = adj[u]; i != -1; i = node[i].next)
		{
			v = node[i].v;
			w = node[i].cost;

			if (d[v] > d[u] + w)
			{
				d[v] = d[u] + w;
				if (!in_q[v])
				{
					in_q[v] = true;
					S.push(v);
				}
			}
		}
	}
}

int main()
{
	int a, b, w;

	scanf("%d%d", &n, &m);

	for (int i = 0; i <= n; i++)
		adj[i] = -1;

	for (int i = 0; i < m; i++)
	{
		scanf("%d%d%d", &a, &b, &w);
		add_edge(a, b, w);
	}

	Dijkstra();

	printf("%d\n", d[n]);
	return 0;
}