Expedition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10025		Accepted: 2918

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. 

* 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

* 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.

4

4 4

5 2

11 5

15 10

25 10

2

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.

挑战上的74页

<span style="font-size:24px;"><a target=_blank href="http://poj.org/searchproblem?field=source&key=USACO+2005+U+S+Open+Gold" style="text-decoration: none;">#include<iostream>

#include<cstdio>

#include<queue>

#include<cstring>

#include<algorithm>

using namespace std;

struct Node{

    int d;

    int v;

    bool operator <(Node b) const{

         return  v<b.v;

    };

}node[10005];

bool cmp(Node a,Node b)

{

    return a.d<b.d;

}

int main()

{

    int n,s,p;

    scanf("%d",&n);

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

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

        scanf("%d %d",&s,&p);

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

                node[i].d=s-node[i].d;

        sort(node+1,node+n+1,cmp);

        priority_queue<Node> q;

        n++;

        node[n].v=0;node[n].d=s;//为了方便，将终点也设成一个加油站

        int t=p,cnt=0;

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

        {

            int x=node[i].d-node[i-1].d;

            while(x>t) /*刚开始是写成if，，wrong 了无数次，后来才发现到达某个点不一定是取一个加油站，可以取多个然后过该点。*/</a></span>

<span style="font-size:24px;"><a target=_blank href="http://poj.org/searchproblem?field=source&key=USACO+2005+U+S+Open+Gold" style="text-decoration: none;">            {

                if(q.empty())

                {

                    printf("-1\n");

                    return 0;

                }

                t+=q.top().v;

                q.pop();

                cnt++;

            }

            q.push(node[i]);/*能到达该点，就能在它这里加油，因此入维护油量最大的优先队列*/

            t-=x;

        }

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

    return 0;

}</a><a target=_blank href="http://poj.org/searchproblem?field=source&key=USACO+2005+U+S+Open+Gold">

</a></span>