代码

好吧，她其实没什么参考价值，可读性极差，而且还最麻烦，来寻求参考价值的人可以不用看了

#include<stdio.h>
#include<math.h>
int main(){
    int i,j,a,b,c,time[5],bagin[5],direction[5],floor,add=0;
    int judt,nowf,add1;
    for(i=0;i<5;i++)
    scanf("%d %d %d",&time[i],&bagin[i],&direction[i]);
    for(i=0;i<5;i++)
        for(j=0;j<4;j++){
            if(time[i]>time[1]){
                a=time[i];
                time[i]=time[i+1];
                time[i+1]=a;
                b=bagin[i];
                bagin[i]=bagin[i+1];
                bagin[i+1]=b;
                c=direction[i];
                direction[i]=direction[i+1];
                direction[i]=c;
            }
            else
            continue;
        }//给他们按请求时间排序
    for(i=0;i<5;i++){
        if(direction[i]==0)
            judt=(bagin[i]-1)+1+(10-bagin[i])+time[i];//judt指乘客到目的地所需时间 
        else
            judt=bagin[i]+1+bagin[i]+time[i];  
        if(judt>=time[i+1]){//判断中途是否两次请求一同执行 
            if(direction[i]==0){//对目的地进行分类，相同则继续判断 
                if((time[i]+bagin[i])>time[i+1])//判断电梯在接受第二次请求时的楼层 
                    nowf=1+time[i+1]-time[i];
                else if((time[i]+bagin[i])==time[i])
                    nowf=bagin[i];
                else 
                    nowf=time[i+1]-(time[i]+bagin[i])+bagin[i];
                if((10-nowf)>abs(nowf-bagin[i+1])){//判断是否一同完成第二次请求 
                    add+=time[i+1]+abs(nowf-bagin[i+1])+1+(10-bagin[i+1]);
                    add1=time[i+1]+abs(nowf-bagin[i+1])+1+2*(10-bagin[i+1]);
                    printf("%d时，停靠在10楼\n%d时，停靠在10楼\n",add1,add1);
                    i++;
                    continue;
                }
                else{
                    add+=judt;
                    printf("%d时，停靠在10楼\n",add);
                    continue;
                }
            }
            else{
                if((time[i+1]-time[i]+2)<=bagin[i]){
                    nowf=time[i+1]-time[i]+1;
                    if(nowf<=bagin[i+1]){
                        add+=time[i+1]-time[i]+2*((bagin[i+1]-nowf+1)+bagin[i]-bagin[i+1]+bagin[i]-1);
                        add1=time[i+1]-time[i]+((bagin[i+1]-nowf+1)+bagin[i]-bagin[i+1]+bagin[i]-1);
                        i++;
                        printf("%d时，停靠在1楼\n%d时，停靠在1楼\n",add1,add1);
                        continue;
                    }
                    else{
                        add+=time[i+1]-time[i]+2*(nowf-bagin[i+1]+bagin[i]-bagin[i+1]+bagin[i]-1);
                        add1=time[i+1]-time[i]+(nowf-bagin[i+1]+bagin[i]-bagin[i+1]+bagin[i]-1);
                        i++;
                        printf("%d时，停靠在1楼\n%d时，停靠在1楼\n",add1,add1);
                        continue;
                 }
                }
                else{
                    nowf=bagin[i]-(bagin[i+1]-(bagin[i]+time[i]+1));
                    if(nowf>=bagin[i+1]){
                        add+=time[i+1]-time[i]+2*(abs(bagin[i+1]-nowf)+bagin[i+1]);
                        add1=time[i+1]-time[i]+(abs(bagin[i+1]-nowf)+bagin[i+1]);
                        i++;
                        printf("%d时，停靠在1楼\n%d时，停靠在1楼\n",add1,add1);
                        continue;
                    }
                    else{
                        add+=time[i]+bagin[i]+bagin[i];
                        printf("%d时，停靠在1楼\n",add);
                        continue;
                    }
                }
            } 
        }
        else{
            if(direction[i]==0){
                judt=(bagin[i]-1)+1+(10-bagin[i])+time[i];
                printf("%d时，停靠在10楼\n",judt);
            }
            else{
                judt=bagin[i]+1+bagin[i]+time[i];
                printf("%d时，停靠在1楼\n",judt);
            }
            add+=judt;
        }
    }
    return 0;
}