Poj 2431 partition dition

Partition dition

Time limit:1000 ms		Memory limit:65536 K
Total submissions:7255		Accepted:2163

Description

A group of cows grabbed a truck and ventured on an unknown dition 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 too tively 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.

Question meaning:A group of cows are driving for adventure. Now the fuel tank is broken. Every time one unit is taken, one unit of oil needs to be consumed, and repair is urgently needed. We need to go to the town located at the current location, there are P units of oil, and there are n fuel points on this road. Now we know the distance between each fuel point and the greatest number of fuel points that can be refueled, select a strategy to minimize the number of refuel stops.

Problem: on the way from a truck to the terminal, fuel can only be refueled at a gas station. However, if you think that "when you arrive at the gas station I, the right to add Bi units of gasoline at any time afterwards is obtained. "The same should be true for solving the problem. When you need to refuel later, you can think that it is the fuel that has been added to the gas station before. Therefore, it seems to be a good method to refuel when the fuel reaches the finish point as few as possible. When the fuel is 0, select the gas station with the largest fuel adding amount Bi;

Note: The input is not in an ordered order and must be sorted first;

#include <iostream>#include <algorithm>#include <queue>using namespace std;int n,l,p;struct sl{int a,b;}s[100005]; bool cmp(sl x,sl y){if (x.a<y.a)return true;return false;}void solve(){s[n].a=l;s[n].b=0;n++;sort(s,s+n,cmp);priority_queue<int> que;int ans=0,pos=0,tank=p;for (int i=0;i<n;i++){int d=s[i].a-pos;while (tank-d<0){if (que.empty()){cout<<-1<<endl;return ;}tank+=que.top();que.pop();ans++;}tank-=d;pos=s[i].a;que.push(s[i].b);}cout<<ans<<endl;}int main(){while (cin>>n){for (int i=0;i<n;i++)cin>>s[i].a>>s[i].b;cin>>l>>p;for(int i=0;i<n;i++)s[i].a=l-s[i].a;solve();}return 0;}

Poj 2431 partition dition