POJ 2431 Expedition (优先队列
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Expedition
Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 13904Accepted: 3940
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
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.
写得第一道 优先队列的题, 结果题目没说那个N的位置是默认排序的 所以要重新排下序 //因为这个wa了好几次AC代码:
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.
写得第一道 优先队列的题, 结果题目没说那个N的位置是默认排序的 所以要重新排下序 //因为这个wa了好几次AC代码:
#include<stdio.h>#include<string.h>#include<queue>#include<stack>#include<algorithm>using namespace std;struct node{ int a,b;}p[10010];bool cmp(node x,node y){ return x.a < y.a;} int main(){ int n, l, pl; while(~scanf("%d",&n)) { int flag = 0; for(int i = 0;i < n; i++) { scanf("%d%d",&p[i].a,&p[i].b); } scanf("%d%d",&l,&pl); for(int i = 0;i < n; i++) p[i].a = l - p[i].a; sort(p,p+n,cmp); p[n].a = l, p[n].b = 0; n++; priority_queue<int> fq; int k = 0, pos = 0; for(int i = 0;i < n; i++) { int d = p[i].a - pos; while(pl - d < 0) { if(fq.empty()) { // printf("-1\n"); flag = 1; break; } pl += fq.top(); fq.pop(); k++; } pl -= d; pos = p[i].a; fq.push(p[i].b); } if(flag) printf("-1\n"); else printf("%d\n",k); }return 0;}
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