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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