POJ 2413 —— 贪心
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Expedition
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. Source
USACO 2005 U S Open Gold
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这道题没啥特别的trick，有一个就是题目输入的加油站可能要手动排一下序。
题意是给你一个长度L，初始汽油P，给你路上的若干的加油站的位置和油量，求到达终点所需的最小加油次数。
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int MOD = (int)1e9 + 7;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int n , l , p;
struct Node
{
int x , y;
}a[MAXN];
bool cmp(Node a , Node b)
{
return a.x < b.x;
}
void init()
{
FOR(i ,0 , n)
{
scanf("%d%d" , &a[i].x , &a[i].y);
}
scanf("%d%d" , &l , &p);
FOR(i ,0 , n)
{
a[i].x = l - a[i].x;
// Bug(a[i])
}
sort(a , a +n , cmp);
// FORR(i ,1 , n)cout << "a[i] = " << a[i] << " b[i] = " << b[i] << endl;
}
void solve()
{
int pos = 0 , ans = 0 ,tank = p;
a[n].x = l , a[n].y = 0;
n++;
// while(!pq.empty())pq.pop();
priority_queue <int> pq;
FOR(i , 0 , n)
{
int d = a[i].x - pos;
// Bug(d)
// cout << "tank = " << tank << endl;
while(tank < d)
{
if(pq.empty())
{
printf("-1\n");
return;
}
tank += pq.top();
pq.pop();
ans++;
}
tank -= d;
pq.push(a[i].y);
pos = a[i].x;
}
printf("%d\n" , ans);
}
int main()
{
// freopen("heritage.in","r",stdin);
// freopen("heritage.out","w",stdout);
while(~scanf("%d" , &n))
{
init();
solve();
}
return 0;
}
/*
*/