传送门:【HDU】1534 Schedule Problem
题目分析:比较简单的差分约束。
每个任务X设一个开始时间Xs,结束时间Xe。
如果任务X需要时间t完成。设不等式:Xe - Xs >= t , Xs - Xe >= -t
SAS X Y:Xs - Ys >= 0
SAF X Y:Xs - Ye >= 0
FAS X Y:Xe - Ys >= 0
FAF X Y:Xe - Ye >= 0
最后将形如Y - X >= d的不等式建边( X , Y , d ),跑一遍spfa最长路,如果无正权环则有解,d[ i ](1 <= i <= n)则为第i个任务的满足条件的最小开始时间。如果有正权环出现则无解。
题目坑的不行啊,一个数据范围都没有给。
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define REP( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )
const int MAXN = 400 ;
const int MAXE = 1000000 ;
const int INF = 0x3f3f3f3f ;
struct Edge {
int v , c ;
Edge* next ;
} E[MAXE] , *H[MAXN] , *cur ;
int d[MAXN] ;
int Q[MAXN] , head , tail ;
bool inq[MAXN] ;
int in[MAXN] ;
int n ;
void init () {
cur = E ;
CLR ( H , 0 ) ;
}
void addedge ( int u , int v , int c ) {
cur -> v = v ;
cur -> c = c ;
cur -> next = H[u] ;
H[u] = cur ++ ;
}
bool spfa () {
CLR ( in , 0 ) ;
CLR ( inq , 0 ) ;
FOR ( i , 0 , n << 1 ) d[i] = -INF ;
head = tail = 0 ;
d[0] = 0 ;
Q[tail ++] = 0 ;
while ( head != tail ) {
int u = Q[head ++] ;
if ( head == MAXN ) head = 0 ;
inq[u] = 0 ;
for ( Edge* e = H[u] ; e ; e = e -> next ) {
int v = e -> v ;
if ( d[v] < d[u] + e -> c ) {
d[v] = d[u] + e -> c ;
if ( !inq[v] ) {
if ( ++ in[v] > 2 * n + 1 ) return 0 ;
inq[v] = 1 ;
Q[tail ++] = v ;
if ( tail == MAXN ) tail = 0 ;
}
}
}
}
return 1 ;
}
void solve () {
int x , y ;
char s[5] ;
init () ;
FOR ( i , 1 , n ) addedge ( 0 , i , 0 ) ;
FOR ( i , 1 , n ) {
scanf ( "%d" , &x ) ;
addedge ( i , i + n , x ) ;
addedge ( i + n , i , -x ) ;
}
while ( ~scanf ( "%s" , s ) && s[0] != '#' ) {
scanf ( "%d%d" , &x , &y ) ;
if ( s[0] == 'S' ) {
if ( s[2] == 'S' ) addedge ( y , x , 0 ) ;
else addedge ( y + n , x , 0 ) ;
} else {
if ( s[2] == 'S' ) addedge ( y , x + n , 0 ) ;
else addedge ( y + n , x + n , 0 ) ;
}
}
if ( spfa () ) FOR ( i , 1 , n ) printf ( "%d %d\n" , i , d[i] ) ;
else printf ( "impossible\n" ) ;
}
int main () {
int cas = 0 ;
while ( ~scanf ( "%d" , &n ) && n ) {
if ( cas ) printf ( "\n" ) ;
printf ( "Case %d:\n" , ++ cas ) ;
solve () ;
}
return 0 ;
}