学步园

Schedule Problem

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Time Limit: 2 Seconds     
Memory Limit: 65536 KB      Special Judge

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A project can be divided into several parts. Each part should be completed continuously. This means if a part should take 3 days, we should use a continuous 3 days do complete it. There are four types of constrains among these parts which are FAS, FAF, SAF
and SAS. A constrain between parts is FAS if the first one should finish after the second one started. FAF is finish after finish. SAF is start after finish, and SAS is start after start. Assume there are enough people involved in the projects, which means
we can do any number of parts concurrently. You are to write a program to give a schedule of a given project, which has the shortest time.

Input

The input file consists a sequences of projects.

Each project consists the following lines:

the count number of parts (one line) (0 for end of input)

times should be taken to complete these parts, each time occupies one line

a list of FAS, FAF, SAF or SAS and two part number indicates a constrain of the two parts

a line only contains a '#' indicates the end of a project

Output

Output should be a list of lines, each line includes a part number and the time it should start. Time should be a non-negative integer, and the start time of first part should be 0. If there is no answer for the problem, you should give a non-line output containing
"impossible".

A blank line should appear following the output for each project.

Sample Input

3
2
3
4
SAF 2 1
FAF 3 2
#
3
1
1
1
SAF 2 1
SAF 3 2
SAF 1 3
#
0

Sample Output

Case 1:
1 0
2 2
3 1

Case 2:

impossible

这题还是挺简单的，对于四种模式，分别建立不同的边,设s[i]表示第i个部分的开始时间，t[i]表示其持续时间

FAS:s[i] + t[i] >= s[j]

FAF:s[i] + t[i] >= s[j] + t[j]

SAF:s[i] >= s[j] + t[j]

SAS:s[i]  >= s[j]

由于题目要求的是the short time，所以只要求最长路就可以了，因图可能不连通，所以加一个超级源点0，到每个点的权值是0

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  

using namespace std;

const int N = 1010;
const int M = 200010;
const int inf = 0x3f3f3f3f;

queue <int>qu; 
int tot, n, m;
int head[N];
int t[N];
int dist[N];
int in_queue[N];

struct node
{
	int weight;
	int next;
	int to;
}edge[M];

void addedge(int from, int to, int weight)
{
	edge[tot].weight = weight;
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

bool spfa(int v0)
{
	memset (dist, -inf, sizeof(dist) );
	memset (in_queue, 0, sizeof(in_queue) );
	dist[v0] = 0;
	in_queue[v0]++;
	while ( !qu.empty() )
	{
		qu.pop();
	}
	qu.push(v0);
	while ( !qu.empty() )
	{
		int u = qu.front();
		qu.pop();
		for (int i = head[u]; ~i; i = edge[i].next)
		{
			int v = edge[i].to;
			if (dist[v] < dist[u] + edge[i].weight)
			{
				dist[v] = dist[u] + edge[i].weight;
				in_queue[v]++;
				if (in_queue[v] == n + 1)
				{
					return false;
				}
				qu.push(v);
			}
		}
	}
	return true;
}

int main()
{
	int icase = 1;
	while (~scanf("%d", &n), n)
	{
		memset (head, -1, sizeof(head) );
		tot = 0;
		char str[10];
		int u, v;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &t[i]);
		}
		while (scanf("%s", str))
		{
			if (str[0] == '#')
			{
				break;
			}
			scanf("%d%d", &u, &v);
			if (!strcmp(str, "FAS"))
			{
				addedge(v, u, -t[u]);
			}
			else if (!strcmp(str, "FAF"))
			{
				addedge(v, u, t[v] - t[u]);
			}
			else if (!strcmp(str, "SAF"))
			{
				addedge(v, u, t[v]);
			}
			else
			{
				addedge(v, u, 0);
			}
		}
		for (int i = 1; i <= n; ++i)
		{
			addedge(0, i, 0);
		}
		printf("Case %d:\n", icase++);
		bool flag = spfa(0);
		if (!flag)
		{
			printf("impossible\n\n");
			continue;
		}
		for (int i = 1; i <= n; ++i)
		{
			printf("%d %d\n", i, dist[i]);
		}
		printf("\n");
	}
	return 0;
}