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CodeForces 709B Checkpoints (数学,最短路)

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)