hdu 1534 Schedule Problem (差分约束)

Schedule Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1085    Accepted Submission(s): 448
Special Judge

Problem 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

 

Source

Asia 1996, Shanghai (Mainland China)

 

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//0MS    248K    1537 B    G++
/*

    题意：
        给出完成作业需要的时间，以及它们间完成的先后关系，问是否可行，可行输出每个作业的开始时间
        
    差分约束：
    
        有n个作业，第 i 个作业所需的时间是 a[i];

        SAS u v 表示 v开始后 u 才能开始；f(u)>=f(v);

        SAF u v 表示 v结束后 u 才能开始；f(u)+a[u]>=f(v);

        FAF u v 表示 v结束后 u 才能结束；f(u)+a[u]>=f(v)+a[v];

        FAS u v 表示 v开始后 u 才能结束；f(u)>=f(v)+a[v] 
 
    这里使用bellman_ford算法，建立反向边，求最长路径 
    
*/
#include<stdio.h>
#include<string.h>
#define N 1005
#define inf 0x7ffffff
struct node{
    int u,v,w;
}edge[4*N];
int d[N];
int a[N];
int n,edgenum;
bool bellman_ford()
{
    memset(d,0,sizeof(d));
    bool flag=true;
    for(int i=0;i<=n;i++){
        if(!flag) break;
        flag=false;
        for(int j=0;j<edgenum;j++){
            if(d[edge[j].v]<d[edge[j].u]+edge[j].w){
                d[edge[j].v]=d[edge[j].u]+edge[j].w;
                flag=true;
            }
        }
    }
    return flag; //如果执行n次后还能松弛证明有正权环 
} 
int main(void)
{
    char opr[5];
    int x,y;
    int k=1;
    while(scanf("%d",&n),n)
    {
        edgenum=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        while(scanf("%s",opr)){
            if(strcmp(opr,"#")==0) break;
            scanf("%d%d",&x,&y);
            edge[edgenum].u=y;
            edge[edgenum].v=x;
            if(strcmp(opr,"SAS")==0){
                edge[edgenum].w=0;
            }
            if(strcmp(opr,"SAF")==0){
                edge[edgenum].w=a[y];
            }
            if(strcmp(opr,"FAS")==0){
                edge[edgenum].w=-a[x];
            }
            if(strcmp(opr,"FAF")==0){
                edge[edgenum].w=a[y]-a[x];
            }
            edgenum++;
        }
        printf("Case %d:
",k++);
        if(bellman_ford()) puts("impossible");
        else{
            for(int i=1;i<=n;i++)
                printf("%d %d
",i,d[i]);
        }
        printf("
");
    }
    return 0;
}

 再贴一个SPFA的：

//218MS    456K    1791 B    G++
#include<iostream>
#include<vector>
#include<queue>
#define N  1005
#define inf 0x7ffffff
using namespace std;
struct node{
    int v,w;
    node(int a,int b){
        v=a;w=b;
    }
};
vector<node>V[N];
int a[N];
int d[N],in[N],vis[N];
int n;
bool spfa()
{
    memset(in,0,sizeof(in));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<=n;i++) d[i]=-inf;
    queue<int>Q;
    Q.push(0);
    vis[0]=1;
    in[0]=1;
    d[0]=0;
    while(!Q.empty()){
        int u=Q.front();
        Q.pop();
        if(in[u]>n) return false;
        vis[u]=0;
        int n0=V[u].size();
        for(int i=0;i<n0;i++){
            int v=V[u][i].v;
            int w=V[u][i].w;
            if(d[v]<d[u]+w){
                d[v]=d[u]+w;
                if(!vis[v]){
                    in[v]++;
                    Q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    return true;
}
int main(void)
{
    string opr;
    int x,y;
    int k=1;
    while(cin>>n)
    {
        if(!n) break;
        for(int i=0;i<=n;i++) V[i].clear();
        for(int i=1;i<=n;i++){
            cin>>a[i];
            V[0].push_back(node(i,0));
        }
        while(cin>>opr){
            if(opr=="#") break;
            cin>>x>>y;
            if(opr=="SAS") V[y].push_back(node(x,0));
            if(opr=="SAF") V[y].push_back(node(x,a[y]));
            if(opr=="FAS") V[y].push_back(node(x,-a[x]));
            if(opr=="FAF") V[y].push_back(node(x,a[y]-a[x]));
        }
        cout<<"Case "<<k++<<":"<<endl;
        if(!spfa()) cout<<"impossible"<<endl;
        else{
            for(int i=1;i<=n;i++)
                cout<<i<<" "<<d[i]<<endl;
        }   
        cout<<endl;
    }
    return 0;
}