poj 3169 Layout 差分约束

       设d[i] 为 第i只牛到第1只牛的距离

       根据题意有：

       d[1] = 0;

       d[b] - d[a] <= D(ml)     

       d[b] – d[a] >= D(md)  

       d[i] – d[i-1] >= 0   题目说牛是从1~n排列下去，这是题目隐藏的条件，刚开始没有加这个条件也能ac

       d[i] – d[1] >= 0    (1<= i <= n)

       以1为源点开始搜索，d[n]就是答案，当d[n] = INF时，说明n是自由点，输出-2

 

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

const int MAX = 50000;
const int INF = 1000000000;
const int N = 1005;

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

Node node[MAX];

int adj[N];
int d[N];
bool in_q[N];
queue<int> Q;
int size;
int n, ml, md;
int cnt[N];

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++;
}

bool spfa()
{
	memset(in_q, false, sizeof(in_q));
	memset(cnt, 0, sizeof(cnt));

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

	d[1] = 0;
	in_q[1] = true;
	Q.push(1);
	int u, v, w;

	while (!Q.empty())
	{
		u = Q.front();
		Q.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;
					Q.push(v);
				}

				if (++cnt[v] > n) return false;
			}
		}
	}

	return true;
}

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

	scanf("%d%d%d", &n, &ml, &md);

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

	size = 0;
	for (int i = 0; i < ml; i++)
	{
		scanf("%d%d%d", &a, &b, &c);
		add_edge(a, b, c); //d[b] - d[a] <= c
	}

	for (int i = 0; i < md; i++)
	{
		scanf("%d%d%d", &a, &b, &c);
		add_edge(b, a, -c);  //d[a] - d[b] <= -c
	}

	for (int i = 2; i <= n; i++)
	{
		add_edge(i, 1, 0); //d[1] - d[i] <= 0
		add_edge(i, i-1, 0); //d[i-1] - d[i] <= 0
	}

	if (spfa())
	{
		if (d[n] == INF)
			printf("-2\n");
		else
			printf("%d\n", d[n]);
	}
	else
		printf("-1\n");
	return 0;
}