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HDU1534 Schedule Problem 【差分约束系统】

2024-07-18 00:23:56   222人阅读

Schedule Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1283    Accepted Submission(s): 534
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) 

题意：给定n个任务的执行时间以及相互之间开始执行时的顺序，求如何安排才能使用时最少。

题解：用SPFA求差分约束系统的最长路，要判断是否有“负环”。


#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 1002
#define maxm maxn * maxn
#define inf 0x3f3f3f3f
using std::queue;

int head[maxn], t[maxn], id;
struct Node{
    int to, w, next;
} E[maxm];
int dist[maxn], out[maxn], in[maxn];
bool vis[maxn];

void addEdge(int u, int v, int w)
{
    E[id].to = v; E[id].w = w;
    E[id].next = head[u]; head[u] = id++;
}

bool SPFA(int n)
{
    int i, u, v, tmp;
    for(i = 0; i <= n; ++i){
        vis[i] = out[i] = 0; dist[i] = -inf;
    }
    u = 0; vis[u] = 1; dist[u] = 0;
    queue<int> Q; Q.push(u);
    while(!Q.empty()){
        u = Q.front(); Q.pop(); vis[u] = 0;
        if(++out[u] > n) return false;
        for(i = head[u]; i != -1; i = E[i].next){
            tmp = dist[u] + E[i].w;
            v = E[i].to;
            if(tmp > dist[v]){
                dist[v] = tmp;
                if(!vis[v]){
                    vis[v] = 1; Q.push(v);
                }
            }
        }
    }
    return true;
}

int main()
{
    int n, cas = 1, u, v, i;
    char str[5];
    while(scanf("%d", &n), n){
        for(i = 1; i <= n; ++i)
            scanf("%d", &t[i]);
        memset(head, -1, sizeof(head)); id = 0;
        while(scanf("%s", str), str[0] != '#'){
            scanf("%d%d", &u, &v);
            if(!strcmp(str, "SAS")) addEdge(v, u, 0);
            else if(!strcmp(str, "SAF")) addEdge(v, u, t[v]);
            else if(!strcmp(str, "FAS")) addEdge(v, u, -t[u]);
            else addEdge(v, u, t[v] - t[u]);
        }
        printf("Case %d:\n", cas++);
        for(i = 1; i <= n; ++i)
            addEdge(0, i, 0);
        if(!SPFA(n)){
            printf("impossible\n\n");
            continue;
        }
        for(i = 1; i <= n; ++i)
            printf("%d %d\n", i, dist[i]);
        printf("\n");
    }
    return 0;
}


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