PAT 1033. To Fill or Not to Fill (25)(贪心)

标签：pat   算法   

其实题目本身不难，难就难在对贪心策略的选取

在每一个加油点应该做这样的判断：
1、首先计算从本加油点做多可能达到的加油点
2、如果有可达加油点的话：

2.1 寻找有无比当前加油点油价便宜的加油点，如果有的话就跑到该便宜的加油点，油量加到支持到该加油点即可
                2.2 如果没有比当前加油点更便宜的加油点的话，又分两种情况：
                2.2.1、如果本次加油可能直接到终点，就加可以到达终点的油量
               2.2.2、否则的话，加满
3、如果没有可达加油点的话：
3.1 看是否可以直接达到终点
3.2 不能达到终点计算最大抵达路程

warning:这里可能没有第0号加油站哦

代码如下：
 

#include<iostream>
#include<fstream>
#include<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;

struct station{
    double distance;
    double price;
};

vector<station>stations;//加油站信息
double minPrice = 99999999;// 最小的价格
double maxArrived = 0;//最大能够抵达的距离

int cmax,d,davg,n;

bool cmp(const station &s1 ,const station &s2){
    if(s1.distance>s2.distance)return false ;
    else return true;
}

/**
@parm currTotalPrice //当前消费
@parm curr 当前位置
@parm gas 当前剩余油量
*/
void find_next_station(double currTotalPrice,int curr,double gasLength,double max_len){
    vector<int>sindex;
    double currLen = (stations[curr]).distance;
    int currIndex = curr;
    double cheapPrice =9999999;
    int cheapIndex=0;
    bool flag = false;
    for(int i=curr+1;i<stations.size();++i){
        if(!flag&&stations[i].distance-currLen <= max_len ){
            sindex.push_back(i);
            if(stations[i].price<stations[curr].price){
                cheapPrice = stations[i].price;
                cheapIndex = i;
                flag=true;
            }

        }
    }
    if(sindex.size()>0){
        if(cheapPrice<stations[curr].price){//下一个最便宜的比当前的便宜
            double pushGasLen = stations[cheapIndex].distance - stations[curr].distance-gasLength;
            currTotalPrice+=stations[curr].price * pushGasLen*1.0/(davg*1.0) ;
            gasLength+= pushGasLen;
        }else{
            //下一个加油站没有比当前便宜的了
            /**
                如果本次可以加油到终点，那就直接加到终点的油量
                否则的话直接加满
            */
            if(stations[curr].distance+max_len>=d){
                double pushGasLen = d - stations[curr].distance -gasLength;
                currTotalPrice += stations[curr].price * pushGasLen*1.0/(davg*1.0);
                if(currTotalPrice < minPrice){
                    minPrice=currTotalPrice;
                }
                maxArrived = d;
                return ;

            }else{
                double pushGasLen = max_len - gasLength;
                currTotalPrice+=stations[curr].price * pushGasLen*1.0/(1.0*davg);
                gasLength = max_len;
            }
        }
        if(cheapIndex!=0)
            curr = cheapIndex;
        else
            curr = sindex.at(sindex.size()-1);
        gasLength -= stations[curr].distance-stations[currIndex].distance;
        find_next_station(currTotalPrice,curr,gasLength,max_len);
    }else{
        /**
            即使加满了也不能或者没有下一个加油站了
            1、直接抵达终点
            2、永远抵达不了终点了
        */
        if(stations[curr].distance+max_len>=d){//可达
            double lastDistance = d-stations[curr].distance;
            currTotalPrice =currTotalPrice+ (lastDistance*1.0/davg)*stations[curr].price;
            if(currTotalPrice<minPrice)
                minPrice = currTotalPrice;
            maxArrived = d;
           // cout<<"minPrice "<<minPrice<<endl;
        }else{//不可达
            double distance = stations[curr].distance+max_len;
            if(distance>maxArrived){
                 maxArrived = distance;
            }
        }
        return ;
    }
}

int main(){
    ifstream cin("data.txt");
    cin>>cmax>>d>>davg>>n;
    const double MAX_LEN=cmax*davg;
    bool noway =true;
    for(int i=0;i<n;++i){
        station s;
        cin>>s.price>>s.distance;
        stations.push_back(s);
        if(s.distance==0){
            noway=false;
        }
    }
    if(noway)//没有第0个加油站，汽车启动不了啊
    {
        printf("The maximum travel distance = %.2lf\n",maxArrived);
        return 0;
    }
    sort(stations.begin(),stations.end(),cmp);
    find_next_station(0,0,0,MAX_LEN);
    if(maxArrived<d){
        printf("The maximum travel distance = %.2lf\n",maxArrived);
    }else{
        printf("%.2lf\n",minPrice);
    }
    return 0;
}

PAT 1033. To Fill or Not to Fill (25)(贪心),布布扣,bubuko.com

PAT 1033. To Fill or Not to Fill (25)(贪心)

标签：pat   算法   

原文地址：http://blog.csdn.net/youmengjiuzhuiba/article/details/38665925