求四人过桥问题的C语言程序

#include <stdio.h>

int time[1000]

int main()

{

int person

int i, j

int index

int sum=0

scanf("%d", &person)

for(i=0i<personi++)

scanf("%d", &time[i])

if(1==person)

{

sum = time[0]

printf("%d\n", sum)

return 0

}

else if(2==person)

{

sum = time[1]

printf("%d\n", sum)

return 0

}

else if(3==person)

{

sum = time[0]+time[1]+time[2]

printf("%d\n", sum)

return 0

}

//从小到大排序

for(i=0i<personi++)

for(j=i+1j<personj++)

{

int tmp

if(time[i]>time[j])

{

tmp = time[i]

time[i] = time[j]

time[j] = tmp

}

}

index = person-1

while(index>2)

{

sum += (time[0]+2*time[1]+time[index])

index -= 2

}

if(2==index)

{

sum += (time[0]+time[1]+time[2])

}

else

{

sum += time[1]

}

printf("%d\n", sum)

return 0

}

设信号量： MUTEX=1 (东西方互斥)

MD=1(东向西使用计数变量互斥)

MX=1(西向东使用计数变量互斥)

设整型变量： CD=0 (东向西的已上桥人数)

CX=0 (西向东的已上桥人数)

从东向西：

P (MD)

IF (CD=0)

{P (MUTEX) }

CD=CD+1

V (MD)

过桥

P (MD)

CD=CD-1

IF (CD=0)

{V (MUTEX) }

V (MD)

从西向东：

P (MX)

IF (CX=0)

{P (MUTEX) }

CX=CX+1

V (MX)

过桥

P (MX)

CX=CX-1

IF (CX=0)

{V (MUTEX) }

V (MX)

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