POJ 2431 Expedition（探险）

POJ 2431 Expedition（探险）

Time Limit: 1000MS   Memory Limit: 65536K

【Description】

【题目描述】

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.

一群奶牛劫了量卡车去丛林探险。可惜不是老司机，偏要赶车上石去作死，最后折了油箱君。现在每行驶一单位路程就会漏一份油。 为了修车，这群奶牛需要经过千难万阻前往最近的城镇（不超过1,000,000个单位路程）。一路上有N(1 <= N <= 10,000)个加油站可供加油（每个加油站有1..100份的油）。 深山老林，人畜皆害。因此，奶牛们希望在去城镇的路上尽可能少地去加油站。幸运的是，车上用的可是四次元油箱，木有装油上限。卡车现在离城镇有L个单位距离，并且有P份的油（1 <= P <= 1,000,000)。 找出到达城镇最少的加油次数，或者这群牛无法抵达城镇。

【Input】	【输入】
* 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	*第1行：一个整数N *第2..N+1行：每行两个以空格隔开的整数表示一个加油站：第一个整数表示城镇到加油站距离；第二个表示加油站的油量。 *第N+2行：两个以空格隔开的整数，L和P

【Output】	【输出】
* 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.	*第1行：一个整数表示到达城镇时最少需要的加油站数量。如果无法达到，输出-1。

【Sample Input - 输入样例】	【Sample Output - 输出样例】
4 4 4 5 2 11 5 15 10 25 10	2

【Hint】

【提示】

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.

输入说明： 卡车离城镇25个单位长度，拥有10个单位汽油。一路上有4个加油站分别离城镇4，5，11和15（因此他们一开始距卡车21，20，14和10）。这些加油站可分别提供4，2，5和10份的汽油。 输出说明： 行驶10个单位长度，获得10份汽油，再行驶4个单位长度，获得5份汽油，然后前往城镇。

【题解】

因为油箱是无限容量的，油量==可以前进的路程。

所以我们可以把问题转化一下：

　　有多少油就可以走多远，已经走过的加油站可以随时隔空加油。

至于怎么选择加油站，每次贪心油量最多的加油站就行了。

【代码 C++】

 #include<cstdio>
 #include <queue>
 #include <algorithm>
 std::priority_queue<int, std::vector<int> > fuel;
 struct Stop{
     int fuel, distant;
 }data[];
 bool cmp(Stop A, Stop B){
     return A.distant < B.distant;
 }
 int main(){
     int i, j, n, L, nowL, opt;
     scanf("%d", &n);
     for (i = ; i < n; ++i){
         scanf("%d%d", &L, &nowL);
         data[i].distant = -L; data[i].fuel = nowL;
     }
     scanf("%d%d", &L, &nowL);
     for (i = ; i < n; ++i) data[i].distant += L;
     std::sort(data, data + n, cmp);

     for (i = opt = ; nowL < L; ++opt){
         while (i < n && data[i].distant <= nowL) fuel.push(data[i++].fuel);
         if (fuel.empty()) break;
         nowL += fuel.top(), fuel.pop();
     }
     if (nowL < L) puts("-1");
     else printf("%d", opt);
     return ;
 }