1033. To-fill or not-to-fill

With highways available, driving a car from Hangzhou to any other city are easy. But since the tank capacity of a car was limited, we have to find gas stations on the the-on-the-the-the-time. Different give Different price. You is 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  (<=), the maximum C Apacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; Davg  (<=20), the average distance per unit gas, 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:pi, the unit gas price, and Di  (<=d), the distance between this station and Hangzhou, for I=1,... N. All the numbers in a line is 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, 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 dist Ance the car can run, accurate up to 2 decimal places.

Tangled up with me all the afternoon and finally it's all kinds of Gaga to figure out

#include <stdio.h> #include <stdlib.h> #include <vector> #include <algorithm>using namespace        Std;typedef struct{float Price;        int dis,period;    }gas;vector<gas> gas;float sum;bool cmp (gas a,gas b); int main () {int cop,distance,per,num,i;    scanf ("%d%d%d%d", &cop,&distance,&per,&num);    Gas temp;         int far=cop*per;//How far up to run for (i=0;i<num;i++) {scanf ("%f%d", &temp.price,&temp.dis);         Gas.push_back (temp);    } sort (Gas.begin (), Gas.end (), CMP);    sum=0.0f;        if (gas[0].dis!=0) {printf ("the maximum travel distance =%.2f\n", sum);        return 0;         } for (i=0;i<num-1;i++) {Gas[i].period=gas[i+1].dis-gas[i].dis;             if (Gas[i].period>far) {Sum+=far;             printf ("The maximum travel distance =%.2f\n", sum);             return 0;         } sum+=gas[i].period; } gas[i].period=distance-gas[i].diS             if (Gas[i].period>far) {Sum+=far;             printf ("The maximum travel distance =%.2f\n", sum);             return 0;    }//Sum+=gas[i].period;    Float cost,min=gas[0].price;//the smallest price int now,left;    Now=gas[0].period;    sum=now;//currently walking the total distance left=far-gas[0].period;//can also walk the distance cost=now*gas[0].price;//at this station spend for (i=1;i<num;i++) {if (gas[i].price<=min)//At this station refueling {min=gas[i].price;//Update minimum price now=gas[i].pe             Riod             Left=far-now;             Cost+=now*gas[i].price; } else {if (gas[i].period<=left)//continue before the refueling {Now+=gas[i].perio                 D                 Left=far-now;                 Cost+=gas[i].period*min;                 } else//start a new refueling {cost+=left*min;                 Min=gas[i].price;                 Now= (Gas[i].period-left);         Cost+=gas[i].price*now;        Left=left-now;    }}} printf ("%.2f\n", cost/per);    System ("pause");    return 0;     }bool cmp (gas A,gas b) {return a.dis<b.dis; }

1033. To-fill or not-to-fill