poj2431

maksyuki 发表于 oj 分类，标签: 基础算法-贪心, 数据结构-队列\双端\单调\优先

01 10月 2015

Expedition

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

5 2

11 5

15 10

25 10

Sample Output

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

题目类型：优先队列+贪心

算法分析：从左往右依次判断卡车是否能够走到下个加油站，如果能，则将当前加油站的可加油量插入到优先队列中(大根堆)。如果不能，则若优先队列为空，直接输出-1，否则优先队列退出一个元素，并加到当前卡车的油量上，重复判断即可。贪心策略是卡车的油能坚持到达下一个加油站的话，就不加。如果加，则尽量拿大油量的加，这样才能保证加的次数尽可能少

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 2e4 + 6;

struct node
{
    int v, w;
};

node aa[maxn];

bool operator < (const node &a, const node &b) {return a.v > b.v;}

int main()
{
//    CFF;
    int n, L, P;
    while (scanf ("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; i++)
            scanf ("%d%d", &aa[i].v, &aa[i].w);

        sort (aa + 1, aa + 1 + n);
        scanf ("%d%d", &L, &P);

        priority_queue <int> qu;
        aa[0].v = L;
        aa[n+1].v = 0;
        int res = 0;
        bool is_valid = true;
        for (int i = 1; i <= n + 1 && is_valid; )
        {
            int dis = aa[i-1].v - aa[i].v;
            if (P >= dis)
            {
                qu.push(aa[i].w);
                P -= dis;
                i++;
            }

            else
            {
                while (P < dis)
                {
                    if (qu.empty())
                    {
                        is_valid = false;
                        break;
                    }

                    int tt = qu.top ();
                    qu.pop ();
                    P += tt;
                    res++;
                }
            }
        }

        if (is_valid)
            printf ("%d\n", res);

        else puts ("-1");
    }
    return 0;
}

100

101

102

103

104

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 2e4 + 6;

struct node

{

int v, w;

};

node aa[maxn];

bool operator < (const node &a, const node &b) {return a.v > b.v;}

int main()

{

// CFF;

int n, L, P;

while (scanf ("%d", &n) != EOF)

{

for (int i = 1; i <= n; i++)

scanf ("%d%d", &aa[i].v, &aa[i].w);

sort (aa + 1, aa + 1 + n);

scanf ("%d%d", &L, &P);

priority_queue <int> qu;

aa[0].v = L;

aa[n+1].v = 0;

int res = 0;

bool is_valid = true;

for (int i = 1; i <= n + 1 && is_valid; )

{

int dis = aa[i-1].v - aa[i].v;

if (P >= dis)

{

qu.push(aa[i].w);

P -= dis;

i++;

}

else

{

while (P < dis)

{

if (qu.empty())

{

is_valid = false;

break;

}

int tt = qu.top ();

qu.pop ();

P += tt;

res++;

}

if (is_valid)

printf ("%d\n", res);

else puts ("-1");

}

return 0;

}

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