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CodeForces 709B Checkpoints

分类讨论。

先对$n$个坐标进行排序。如果$a$在$x[1]$左边,那么肯定是$x[n]$不走。如果$a$在$x[n]$右边,那么$x[1]$不走。

剩下的讨论一下是$x[1]$不走还是$x[n]$不走,几种情况都算一下取最小值即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){    freopen("D:\\in.txt","r",stdin);    freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){    char c = getchar(); x = 0;while(!isdigit(c)) c = getchar();    while(isdigit(c)) { x = x * 10 + c - 0; c = getchar();  }}LL INF=1e17;const int maxn=100010;LL x[maxn],a;int n;int main(){    scanf("%d%lld",&n,&a);    for(int i=1;i<=n;i++) scanf("%lld",&x[i]);    sort(x+1,x+1+n);        if(n==1) {printf("0\n");return 0;}        if(x[1]>=a) { printf("%lld\n",x[n-1]-a); return 0;}    if(a>=x[n]) { printf("%lld\n",a-x[2]); return 0;}    LL f1=INF,f2=INF,f3=INF,f4=INF;    if(x[2]<=a) f1=2*(a-x[2])+(x[n]-a);    else f1=x[n]-a;    f2=2*(x[n]-a)+(a-x[2]);    if(a<=x[n-1]) f3=2*(x[n-1]-a)+(a-x[1]);    else f3=a-x[1];    if(x[1]<=a) f4=2*(a-x[1])+(x[n-1]-a);    printf("%lld\n",min(min(f1,f2),min(f3,f4)));    return 0;}

 

CodeForces 709B Checkpoints