假的POJ#2431 Expedition
Expedition
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 25813 | Accepted: 7160 |
Description
A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every unit of distance it travels.
To repair the truck, the cows need to drive to the nearest town (no more than 1,000,000 units distant) down a long, winding road. On this road, between the town and the current location of the truck, there are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire additional fuel (1..100 units at each stop).
The jungle is a dangerous place for humans and is especially dangerous for cows. Therefore, the cows want to make the minimum possible number of stops for fuel on the way to the town. Fortunately, the capacity of the fuel tank on their truck is so large that there is effectively no limit to the amount of fuel it can hold. The truck is currently L units away from the town and has P units of fuel (1 <= P <= 1,000,000).
Determine the minimum number of stops needed to reach the town, or if the cows cannot reach the town at all.
To repair the truck, the cows need to drive to the nearest town (no more than 1,000,000 units distant) down a long, winding road. On this road, between the town and the current location of the truck, there are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire additional fuel (1..100 units at each stop).
The jungle is a dangerous place for humans and is especially dangerous for cows. Therefore, the cows want to make the minimum possible number of stops for fuel on the way to the town. Fortunately, the capacity of the fuel tank on their truck is so large that there is effectively no limit to the amount of fuel it can hold. The truck is currently L units away from the town and has P units of fuel (1 <= P <= 1,000,000).
Determine the minimum number of stops needed to reach the town, or if the cows cannot reach the town at all.
Input
* Line 1: A single integer, N
* Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the town to the stop; the second is the amount of fuel available at that stop.
* Line N+2: Two space-separated integers, L and P
* Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the town to the stop; the second is the amount of fuel available at that stop.
* Line N+2: Two space-separated integers, L and P
Output
* Line 1: A single integer giving the minimum number of fuel stops necessary to reach the town. If it is not possible to reach the town, output -1.
Sample Input
4 4 4 5 2 11 5 15 10 25 10
Sample Output
2
Hint
INPUT DETAILS:
The truck is 25 units away from the town; the truck has 10 units of fuel. Along the road, there are 4 fuel stops at distances 4, 5, 11, and 15 from the town (so these are initially at distances 21, 20, 14, and 10 from the truck). These fuel stops can supply up to 4, 2, 5, and 10 units of fuel, respectively.
OUTPUT DETAILS:
Drive 10 units, stop to acquire 10 more units of fuel, drive 4 more units, stop to acquire 5 more units of fuel, then drive to the town.
The truck is 25 units away from the town; the truck has 10 units of fuel. Along the road, there are 4 fuel stops at distances 4, 5, 11, and 15 from the town (so these are initially at distances 21, 20, 14, and 10 from the truck). These fuel stops can supply up to 4, 2, 5, and 10 units of fuel, respectively.
OUTPUT DETAILS:
Drive 10 units, stop to acquire 10 more units of fuel, drive 4 more units, stop to acquire 5 more units of fuel, then drive to the town.
Source
题意:一辆卡车从起点驶至距离为L处的town,出发时有P单位油。沿途有N个加油站,每个加油站距离起点(原题中应为距离终点town)Ai单位,有Bi单位油。求到达终点需要的最小加油次数。
思路:每经过一个加油站i即表示获得了加Bi油的权力。每次没油时从所有权力中选出最大的加油,若无权力加油,则表示无法到达终点。
1 #include<iostream> 2 #include<queue> 3 using namespace std; 4 int pos, dis, ans; 5 int A[10010], B[10010]; 6 int main() { 7 int N, L, P; //起初有P单位油 8 cin >> N >> L >> P; 9 for (int i = 0; i < N; i++) { 10 cin >> A[i]; 11 } 12 for (int i = 0; i < N; i++) { 13 cin >> B[i]; 14 } 15 A[N] = L; 16 B[N] = 0; 17 priority_queue<int>que; 18 for (int i = 0; i <= N; i++) { 19 dis = A[i] - pos; 20 P -= dis; //剩余油量 21 while (P < 0) { 22 if (que.empty()) { //无油可加 23 cout << "-1"; 24 cout << ans; 25 return 0; 26 } 27 else { 28 P += que.top(); //加油 29 que.pop(); 30 ans++; 31 } 32 } 33 pos = A[i]; //到达下一点 34 que.push(B[i]); 35 } 36 cout << ans; 37 }