hdu 1534(差分约束)
Schedule Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1715 Accepted Submission(s): 757
Special Judge
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.
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
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.
2
3
4
SAF 2 1
FAF 3 2
#
3
1
1
1
SAF 2 1
SAF 3 2
SAF 1 3
#
0
1 0
2 2
3 1
Case 2:
impossible
#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const int INF = ;
const int N = ;
struct Edge{
int v,w,next;
}edge[];
int head[N];
int n,tot;
int val[N];
void init(){
memset(head,-,sizeof(head));
tot = ;
}
void addEdge(int u,int v,int w,int &k){
edge[k].v = v,edge[k].w = w,edge[k].next = head[u],head[u] = k++;
}
int low[N],time[N];
bool vis[N];
int spfa(int s){
for(int i=;i<=n;i++){
vis[i] = false;
low[i] = -INF;
time[i] = ;
}
low[s] = ;
time[s]++;
queue<int> q;
q.push(s);
int num = ((int)sqrt(n)+); ///改成根号 n 可以AC...
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = false;
for(int k=head[u];k!=-;k=edge[k].next){
int v = edge[k].v,w=edge[k].w;
if(low[v]<low[u]+w){
low[v] = low[u]+w;
if(!vis[v]){
vis[v] = true;
q.push(v);
if(time[v]++>num) return ;
}
}
}
}
return ;
}
int main(){
int t = ;
while(scanf("%d",&n)!=EOF,n){
init();
int MAX = -;
for(int i=;i<=n;i++){
scanf("%d",&val[i]);
}
char str[];
int super = ;
while(scanf("%s",str)){
if(strcmp(str,"#")==) break;
int a,b;
scanf("%d%d",&a,&b);
if(strcmp(str,"SAF")==){
addEdge(b,a,val[b],tot);
}else if(strcmp(str,"FAF")==){
addEdge(b,a,-(val[a]-val[b]),tot);
}else if(strcmp(str,"FAS")==){
addEdge(b,a,-val[a],tot);
}else{
addEdge(b,a,,tot);
}
}
for(int i=;i<=n;i++){
addEdge(super,i,,tot);
}
printf("Case %d:\n",t++);
if(spfa(super)){
for(int i=;i<=n;i++){
printf("%d %d\n",i,low[i]);
}
}else{
printf("impossible\n");
}
printf("\n");
}
return ;
}
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