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;

}