POJ 2674 Linear world （经典题目 -- 这个有坑= =）

题意：

在一条线上，告诉你每个蚂蚁的位置和方向，两只蚂蚁碰面后原路返回，告诉你线的长度，和蚂蚁速度（所有的蚂蚁速度都一样），求最后一只蚂蚁掉落的时间和名字。

思路：

很久之前做的一道思路题目了。 这个题比较坑把 记录一下。

两只蚂蚁碰面后相当于不回头一直往前走。

那么我们可以记录下来每只蚂蚁假如不碰面掉落的时间，取个最大值就是最后一只蚂蚁掉落的时间，在把那个时间对应的每只蚂蚁位置记录下来，还在线上的就是最后一只蚂蚁。

坑：

竟然是截断到两位小数，不是四舍五入，那么直接给ans×100，向下取整后在除以100即可。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <cstdlib>#include <iostream>#include <vector>#define Siz(x) (int)x.size()using namespace std;const int maxn = 320000 + 7;const double eps = 1e-10;int dcmp(double a,double b){    if (fabs(a-b) < eps) return 0;    if (a > b) return 1;    return -1;}char s[320];struct Node{    char name[320];    double pos;int dir;    void read(){        scanf("%s",s);        if (s[0] == 'p' || s[0] == 'P')dir = 1;        else dir = -1;        scanf("%lf",&pos);        scanf("%s",name);    }}p[maxn];vector<double>v;int main(){    double L,t;    int n;    while(~scanf("%d",&n) && n){        scanf("%lf %lf",&L, &t);        v.clear();        double Maxtime = -1;        for (int i = 0; i < n; ++i){            p[i].read();            double pos = p[i].pos;            int dir = p[i].dir;            if (dir == 1){                if ((L-pos)/t > Maxtime)Maxtime = (L-pos)/t;            }            else {                if (pos/t >Maxtime)Maxtime=pos/t;            }        }            int ans;        for (int i = 0; i < n; ++i){            double pos = p[i].pos;            int dir = p[i].dir;            double p2 = pos+dir*t*Maxtime;            v.push_back(p2);        }        sort(v.begin(),v.end());        for (int i = 0; i < Siz(v); ++i){            double p2 = v[i];            if (dcmp(p2,0) >= 0 && dcmp(p2,L) <= 0){                ans= i;                break;            }        }        Maxtime*=100;        Maxtime = (int)Maxtime *1.0;        Maxtime/=100;        printf("%13.2f %s\n",Maxtime,p[ans].name);    }    return 0;}/**113.5 2p 3.5 Smarty410  1p  1  Helgan 3 Joannap  5  Venusn  7  Clever0         5.00**/

Linear world

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3633 Accepted: 813

Description

The Disc, being flat, has no real horizon. Any adventurous sailors who get funny ideas from staring at eggs and oranges for too long and set out for the antipodes soon learned that the reason why distant ships sometimes looked as though they were disappearing over the edge of the world was that they were disappearing over the edge of the world. (Terry Pratchett -Colour of Magic) 
Not so long time ago people used to believe that they live on 2-D world and if they will travel long enough in one direction, they will fall down over the edge. Even when it was proved that the Earth is rounded some of them were still afraid to travel to the southern hemisphere. 
Try to imagine one 1-D (linear) world. On such world there are only two possible directions (left and right). All inhabitants of such world were created exactly at the same time and suddenly all of them start to move (all with same constant velocity) in one or the other direction. If two inhabitants encounter each other, they politely exchange greetings and then they turn around and start to move in an opposite direction. When an inhabitant reaches the end of the world he falls away and disappears. 
Your task is to determine, for a given scenario of creation, which inhabitant and when (counting from the moment of creation) will be the last one to fall away. You can assume that the time required to exchange greetings and turn around is 0.

Input

The input consists of multiple descriptions (data sets) of the creation moment. File structure is as follows: 
N 
LV 
DIR POS NAME 
... 
The first line defines the number of inhabitants (N<32000). Data set starting with value N=0 represents the end of the input file. The second line contains length of the world L(float) and velocity of inhabitants V(float). Both values are always positive. In next N lines the data about inhabitants are given in an order of increasing POS (positive direction): 
DIR – initial direction ('p' or 'P' for positive and 'n' or 'N' for negative) 
POS – position in the time of creation (0<=POS<=L) 
NAME – name of inhabitant (string up to 250 characters) 
Input values within one line are separated with at least one space and there will be no empty lines in input. You may assume that input is always correct and that each data set has only one unique solution.

Output

The output consists of one line per each input data set. The first value should be the time when the last inhabitant will fall of the linear world counting from the moment of creation. Value should be printed truncated to two decimal places in a field 13 characters wide. The second value should be the name of the inhabitant. Values should be separated with single space character.

Sample Input

1   13.5 2   p 3.5 Smarty  4  10  1  p  1  Helga  n 3 Joanna  p  5  Venus  n  7  Clever  0 

Sample Output

         5.00 Smarty         9.00 Venus

Source

Southeastern Europe 2005

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