POJ 2431 Expedition 优先队列

由于油箱是无限的，所以每次经过加油站都不选择加油，等到后面油不够了，再来“反悔”，把前面经过的油站的油加上，尽可能选量大的,用优先队列维护。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
    //freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
const int maxn=10005;
pair<int,int> da[maxn];
int n,L,P;
void solve()
{

    priority_queue<int> q;
    int now=0,remain=P,cnt=0;
    int flag=0;
    for(int i=0;i<n;i++)
    {
        if(remain>=L-now){flag=1;break;}
        while(remain<da[i].first-now)
        {
            if(!q.empty())
            {
                int x=q.top();q.pop();
                remain+=x;
                cnt++;
            }else {printf("-1
"); return ;}
        }
        remain-=(da[i].first-now);
        now=da[i].first;
        q.push(da[i].second);
    }
    while(remain<L-now&&!q.empty())
    {
        remain+=q.top();q.pop();
        cnt++;
    }
    if(remain>=L-now) flag=1;
    printf("%d
",flag?cnt:-1);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            int a,b;scanf("%d%d",&a,&b);
            da[i]=make_pair(a,b);
        }

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

        for(int i=0;i<n;i++)
            da[i].first=L-da[i].first;
        sort(da,da+n);
        solve();

    }
    return 0;
}