Expedition
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7707 | Accepted: 2263 |
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
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.
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff; //无限大
int a[maxn];
int b[maxn];
int c[maxn];
struct fox
{
int a;
int b;
};
fox m[maxn];
bool cmp(fox c,fox d)
{
return c.a<d.a;
}
int main()
{
int n;
while(cin>>n){
for(int i=0;i<n;i++)
{
cin>>m[i].a>>m[i].b;
}
int d,l;
cin>>d>>l;
for(int i=0;i<n;i++)
{
m[i].a=d-m[i].a;
}
sort(m,m+n,cmp);
int pos=0;
int flag=0;
int ans=0;
priority_queue<int> q;
m[n].a=d;
m[n].b=0;
for(int i=0;i<=n;i++)
{
int dis=m[i].a-pos;
if(dis>l)
{
while(1)
{
if(q.empty())
{
flag=1;
break;
}
else
{
l+=q.top();
q.pop();
ans++;
}
if(l>=dis)
break;
}
if(flag==1)
break;
}
l-=dis;
pos=m[i].a;
q.push(m[i].b);
}
if(flag==1)
cout<<"-1"<<endl;
else
cout<<ans<<endl;
}
}