POJ 3618 Exploration

简单题。。。但要注意理解题意，每次都从都选离原点最近的未访问过的点走。。

思路：将-100,000 ≤ x_i ≤ 100,000范围转换成 0 ~ 200,000 OFFSET（偏移量） 100000 ,由于我是用bool visit[N]。。。虽然代码有点长。。但跑出来的效率比较高。。

//376K  63MS
#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <climits>
#define MIN(X,Y) ((X)<(Y)?(X):(Y))
#define MAX(X,Y) ((X)>(Y)?(X):(Y))
#define N  200005
#define OFFSET 100000       //作为偏移量
using namespace std ;bool visit[N] ;void Exploration ( const double t )
{double count_time ;count_time = 0 ;int ans ;           //记录参观的景点数ans = 0 ;int index ;index = OFFSET ;while ( count_time <= t ){int mark1 , mark2 ;     //每次从100000向左，右搜索第一个未遍历过的点int i , k ;for (  i = OFFSET , k = OFFSET ; i < N || k >= 0 ; i ++ , k -- ){if ( visit[i] ){mark1 = i ;break ;}if ( visit[k] ){mark2 = k ;break ;}}double dis1 , dis2 ;dis1 = fabs ( (double)OFFSET - (double)mark1 ) ;dis2 = fabs ( (double)OFFSET - (double)mark2 ) ;int min ;if ( dis1 < dis2 ){min = mark1 ;}else{min = mark2 ;}if ( count_time + fabs( (double)min - index ) > t ){break ;}ans ++ ;count_time  += fabs ( (double)min - index ) ;       //记录时间index = min ;                       //新参观的landmarkvisit[index] = false ;mark1 = mark2 = INT_MAX ;               //注意重置}printf ("%d\n" , ans ) ;
}int
main ( )
{double t ;int n ;while ( EOF != scanf ("%lf%d" , & t , & n  ) ){memset ( visit , 0 , sizeof ( visit ) ) ;for ( int i = 1 ; i <= n ; i ++ ){int index ;scanf ("%d" , & index ) ;index += OFFSET ;visit[index] = true ;}Exploration ( t ) ;}return 0 ;
}

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