Hdu1534 schedule problem [difference constraint system]

Schedule problemtime limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others) Total submission (s): 1283 accepted submission (s): 534 Special Judge
Problem descriptiona project can be divided into several parts. each part shoshould be completed continuously. this means if a part shocould take 3 days, we shocould 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 shoshould 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.
Inputthe 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 shoshould 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
 
Outputoutput shocould be a list of lines, each line should des a part number and the time it shocould start. time shocould be a non-negative integer, And the start time of first part shocould be 0. if there is no answer for the problem, you shoshould give a non-line output containing "impossible ".

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

 
Sample Input

3234SAF 2 1FAF 3 2#3111SAF 2 1SAF 3 2SAF 1 3#0

 
Sample output

Case 1:1 02 23 1Case 2:impossible

 
Sourceasia 1996, Shanghai. Question: If spfa is used to calculate the longest path of the differential constraint system, it is necessary to determine whether there is a "negative ring ".

#include <stdio.h>#include <string.h>#include <queue>#define maxn 1002#define maxm maxn * maxn#define inf 0x3f3f3f3fusing 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;}