zoj 1455 Schedule Problem 差分约束

        设d[i]为 第i个工作开始的时间，根据题意有：

        FAS a b: d[a] + time[a] >= d[b]

        FAF a b: d[a] + time[a] >= d[b] + time[b]

        SAF a b: d[a] >= d[b] + time[b]

        SAS a b: d[a] >= d[b]

       还有： d[i] >= 0 即 d[i] – d[0] >= 0

       以0为源点开始搜索，0是超源点

 

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

const int MAX = 100000;
const int N = 10000;
const int INF = 1000000000;

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

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

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()
{
	queue<int> Q;

	memset(cnt, 0, sizeof(cnt));
	memset(in_q, false, sizeof(in_q));
	for (int i = 0; i <= n; i++)
		d[i] = -INF;

	d[0] = 0;
	Q.push(0);
	in_q[0] = true;
	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()
{
	char c[20];
	int a, b;
	int cases = 1;

	while (scanf("%d", &n) != EOF)
	{
		if (n == 0) break;
		
		
		size = 0;
		for (int i = 0; i <= n; i++)
			adj[i] = -1;

		for (int i = 1; i <= n; i++)
			scanf("%d", &times[i]);

		while (1)
		{
			getchar();
			scanf("%s", c);
			if (strcmp(c, "#") == 0)
				break;
			else if (strcmp(c, "FAS") == 0)
			{
				scanf("%d%d", &a, &b);
				add_edge(b,a, -times[a]);
			}
			else if (strcmp(c, "FAF") == 0)
			{
				scanf("%d%d", &a, &b);
				add_edge(b,a, times[b]-times[a]);
			}
			else if (strcmp(c, "SAF") == 0)
			{
				scanf("%d%d", &a, &b);
				add_edge(b, a, times[b]);
			}
			else if (strcmp(c, "SAS") == 0)
			{
				scanf("%d%d", &a, &b);
				add_edge(b, a, 0);
			}
		}

		for (int i = 1; i <= n; i++)
			add_edge(0, i, 0);
		printf("Case %d:\n", cases++);

		if (spfa())
		{		
			for (int i = 1; i <= n; i++)
				printf("%d %d\n", i, d[i]);
		}
		else	
			printf("impossible\n");
		
		printf("\n");
		
	}
	return 0;
}