1033. To Fill or Not to Fill (25) 贪心算法

1033. To Fill or Not to Fill (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

ZHANG, Guochuan

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive numbers: C_max (<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; D_avg (<=20), the average distance per unit gas that the car can run; and N (<= 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: P_i, the unit gas price, and D_i (<=D), the distance between this station and Hangzhou, for i=1,...N. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print "The maximum travel distance = X" where X is the maximum possible distance the car can run, accurate up to 2 decimal places.

Sample Input 1:

50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 300

Sample Output 1:

749.17

Sample Input 2:

50 1300 12 27.10 07.00 600

Sample Output 2:

The maximum travel distance = 1200.00

思路：输入后将加油站按距离排序。题目假设开始没有油，所以如果没有距离为0的加油站则直接输出0。从当前加油站往前找比当前价格低的加油站，如果找到则加上刚好到那个加油站的油，如果没有，则把油箱加满，到能走到的价格最低的加油站。用cnt检查是否能走到下一个加油站

#include<cstdio> #include<algorithm>using namespace std;typedef struct nod{double pr;double dis;}nod;nod nods[500];double cmax,d,davg;int n;bool cmp(nod a,nod b){return a.dis<b.dis;}int main(){scanf("%lf %lf %lf %d",&cmax,&d,&davg,&n);for(int i=0;i<n;i++){scanf("%lf %lf",&nods[i].pr,&nods[i].dis);}nods[n].dis=d;nods[n].pr=0.0;sort(nods,nods+n+1,cmp);double ttcost=0.0,lefto=0.0;double maxdis=cmax*davg;double nowdis=0.0;int dangqianzhan=0,next;if(nods[0].dis==0.0){    while(nowdis<d){    double nowp=nods[dangqianzhan].pr;    next=9999;    int cnt=0;    for(int i=dangqianzhan+1;i<=n&&nods[i].dis-nowdis<=maxdis&&nods[i].dis<=d;i++){    cnt++;    if(nods[i].pr<nowp){    next=i;    break;}}if(cnt==0) break;if(next==9999){double minp=99999999.0;for(int i=dangqianzhan+1;i<=n&&nods[i].dis-nowdis<=maxdis&&nods[i].dis<=d;i++){    if(nods[i].pr<minp){    minp=nods[i].pr;    next=i;}}ttcost+=(cmax-lefto)*nods[dangqianzhan].pr;lefto=cmax-(nods[next].dis-nowdis)/davg;dangqianzhan=next;nowdis=nods[dangqianzhan].dis;}else{double needgo=nods[next].dis-nowdis;double needo=needgo/davg;ttcost+=(needo-lefto)*nods[dangqianzhan].pr;lefto=0.0;nowdis=nods[next].dis;dangqianzhan=next;}    }    if(nowdis==d)printf("%.2lf",ttcost);    else printf("The maximum travel distance = %.2lf",nowdis+maxdis);    }    else printf("The maximum travel distance = 0.00");    return 0;}