HDU 1534 Schedule Problem（差分约束）

最新推荐文章于 2019-08-14 20:39:58 发布

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最新推荐文章于 2019-08-14 20:39:58 发布

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分类专栏：图论-最短路

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本文介绍了一种使用SPFA算法解决项目调度问题的方法，具体地，通过差分约束系统来确定项目的最短完成时间。文章详细展示了如何根据四种不同类型的任务约束（FAS、FAF、SAF和SAS）构建图，并利用SPFA算法寻找从源节点到各任务节点的最长路径，从而得出每个任务的最早开始时间。

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思路：一道差分约束最长路求最小值，读懂题目连边求一次最长路就可以了

#include<bits/stdc++.h>
using namespace std;
const int maxn = 10010;
#define inf 1e9
vector<pair<int,int> >e[maxn];
int d[maxn],inq[maxn],a[maxn],cnt[maxn];
int n;
int spfa()
{
	memset(inq,0,sizeof(inq));
	memset(cnt,0,sizeof(cnt));
	for(int i = 0;i<=n;i++)
		d[i]=-inf;
	queue<int>q;
	q.push(0);
	inq[0]=1;
	d[0]=0;
    while(!q.empty())
	{
		int u = q.front();
		q.pop();
		inq[u]=0;
		for(int i = 0;i<e[u].size();i++)
		{
			int v = e[u][i].first;
			if(d[v]<d[u]+e[u][i].second)
			{
				d[v]=d[u]+e[u][i].second;
				if(!inq[v])
				{
					inq[v]=1;
					q.push(v);
					if(++cnt[v]>n)
						return false;
				}
			}
		}
	}
	return true;
}
int main()
{
	int cas = 1;
    while(scanf("%d",&n)!=EOF && n)
	{
		for(int i = 0;i<=n;i++)
			e[i].clear();
		for(int i = 1;i<=n;i++)
			scanf("%d",&a[i]);
		char s[105];
		while(scanf("%s",s) && strcmp(s,"#")!=0)
		{
            int u,v;
			scanf("%d%d",&u,&v);
			if(strcmp(s,"FAS")==0)
				e[v].push_back(make_pair(u,-a[u]));
			if(strcmp(s,"FAF")==0)
				e[v].push_back(make_pair(u,a[v]-a[u]));
			if(strcmp(s,"SAF")==0)
				e[v].push_back(make_pair(u,a[v]));
			if(strcmp(s,"SAS")==0)
				e[v].push_back(make_pair(u,0));
		}
		for(int i = 1;i<=n;i++)
			e[0].push_back(make_pair(i,0));
		printf("Case %d:\n",cas++);
		if(spfa())
		{
			for(int i = 1;i<=n;i++)
				printf("%d %d\n",i,d[i]);
		}
		else
			printf("impossible\n");
		printf("\n");
	}
}

Description

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