【CF1020A】New Building for SIS（签到）

题意：

有n栋楼，从一栋楼某个地方，到大另一栋楼的某个地方，每栋楼给了连接楼的天桥，每走一层或者穿个一栋楼花费一分钟，求出起点到大目的点最少花费的时间

n,h<=1e8,q<=1e4

思路：分类讨论

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   1100000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void swap(int &x,int &y)
 {
     int t=x;x=y;y=t;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     int n,h,a,b,k;
     scanf("%d%d%d%d%d",&n,&h,&a,&b,&k);
     for(int i=;i<=k;i++)
     {
         int x1,y1,x2,y2;
         scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
         int ans=<<;
         ans=min(ans,abs(a-y1)+abs(x1-x2)+abs(a-y2));
         ans=min(ans,abs(b-y1)+abs(x1-x2)+abs(b-y2));
         if(a<=y1&&y1<=b) ans=min(ans,abs(x2-x1)+abs(y1-y2));
         if(a<=y2&&y2<=b) ans=min(ans,abs(x2-x1)+abs(y1-y2));
         if(x1==x2) ans=min(ans,abs(y1-y2));
         printf("%d\n",ans);
     }