【PAT Advanced Level】1033. To Fill or Not to Fill (25)

网上策略说的很详细了，利用贪心算法：

假设我们现在在A点，A能到达的最远距离之内有B, C, D三个点：

1. 如果B， C， D中有比A小的，则找到离A最近的，A加的油刚好够到这个点；

2. 如果没有比A小的，A加满油，到B\C\D中最小的一个点停下来，再加油。

代码注意细节：当第一个点距离不是0时，车子哪里都去不了！

#include <iostream>#include <vector>#include <algorithm>#include <fstream>#include <iomanip>using namespace std;double a, b, c, d;struct node{double price;double dist;node(double p, double d): price(p), dist(d) {}};vector<node> v;bool finished = false;double cost = 0;bool cmp(const node &n1, const node &n2){return n1.dist < n2.dist;}double fillOrNot(int index, double left){double fastest = v[index].dist + c * a;if(index + 1 < d && (v[index + 1].dist - v[index].dist) > c * a){cost += v[index].price * (a - left);return fastest;}else if(index + 1 == d && (b - v[index].dist) > c * a){cost += v[index].price * (a - left);return fastest;}/*if(fastest >= b){cost += ((b - v[index].dist)/c - left) * v[index].price;finished = true;return 0.0;}*/int smallest = index + 1;for(int i = index + 1; i < d && v[i].dist < fastest; i++)if(v[index].price > v[i].price){cost += ((v[i].dist - v[index].dist)/c - left)*v[index].price;return fillOrNot(i, 0);}else if(v[smallest].price > v[i].price)smallest = i;//如果没有比当前便宜的if(fastest >= b){//如果当前可以到终点，直接到cost += ((b - v[index].dist)/c - left) * v[index].price;finished = true;return 0.0;}else{cost += (a - left) * v[index].price;return fillOrNot(smallest, a - (v[smallest].dist - v[index].dist)/c);}}int main(){//fstream cin("a.txt");cin>>a>>b>>c>>d;double tmp1;double tmp2;for(int i = 0; i < d; i++){cin>>tmp1>>tmp2;node n(tmp1, tmp2);v.push_back(n);}sort(v.begin(), v.end(), cmp);double fastest = fillOrNot(0, 0.0);if(v[0].dist != 0){cout<<"The maximum travel distance = 0.00"<<endl;return 0;}if(finished)cout<<fixed<<setprecision(2)<<cost<<endl;elsecout<<"The maximum travel distance = "<<fixed<<setprecision(2)<<fastest<<endl;}