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题意：给定你的坐标，和 n 个点，问你去访问至少n-1个点的最短路是多少。
析：也是一个很简单的题，肯定是访问n-1个啊，那么就考虑从你的位置出发，向左访问和向右访问总共是n-1个，也就是说你必须从1 - n-1 全访问一次，
或者是2 - n 全访问一次，有一段是访问了两次，加上就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e8;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int main(){
while(scanf("%d %d", &n, &m) == 2){
for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
if(n == 1){ printf("0\n"); continue; }
sort(a, a+n);
LL ans1 = (LL)a[n-1]-a[1] + Min(abs(m-a[1]), abs(m-a[n-1]));
LL ans2 = (LL)a[n-2]-a[0] + Min(abs(m-a[0]), abs(m-a[n-2]));
printf("%I64d\n", Min(ans1, ans2));
}
return 0;
}
CodeForces 709B Checkpoints (数学，最短路)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5808818.html
