Expedition
Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 8434Accepted: 2474
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.
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
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
44 45 211 515 1025 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.
题意：卡车出发到L距离处 本身有P单位汽油行驶每单位距离耗油1单位汽油耗尽则无法行驶途中有N个加油站第i个加油站距离起点Ai可以给卡车加油Bi 燃料箱无限大的情况下卡车能否抵达终点能抵达的话输出最少需要加多少次油否则输出-1
题解：讲经过每一个加油站看做获得一个加油的机会油耗尽的时候则考虑之前经过的加油站加过油因为要求尽量少所以每次尽可能选加油量Bi最大的加油站优先队列处理使用STL的priority_queue
#include <iostream>#include <cstdio>#include <queue>#include <algorithm>using namespace std;const int maxn = 10000 + 11;struct node{int a, b;}no[maxn];bool cmp(node x, node y){return x.a < y.a;}int main(){int l, p, n;scanf("%d", &n);for(int i = 0; i < n; ++i){scanf("%d%d", &no[i].a, &no[i].b);}scanf("%d%d", &l, &p);for(int i = 0; i < n; ++i){no[i].a = l – no[i].a;}sort(no, no+n, cmp);no[n].a = l;no[n].b = 0;++n;priority_queue<int> q;int ans = 0, pos = 0, tank = p;for(int i = 0; i < n; ++i){int d = no[i].a – pos;while(tank – d < 0){if(q.empty()){puts("-1");return 0;}tank += q.top();q.pop();++ans;}tank -= d;pos = no[i].a;q.push(no[i].b);}printf("%d\n", ans);return 0;}
，坚硬的城市里没有柔软的爱情，生活不是林黛玉，
