学步园

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 ;
}