汽车加油行驶问题

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/*
n个加油站,,邮箱容量不限,,,,每个加油站可加的油量有限,,,求最少的加油次数
因为邮箱容量不限,,,可以这样贪心,,,如果邮箱的油能到达下一站,,则直接到达下一站,,否则,,,从已经走过的加油站中选择油量最多的站加一次油
*/
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
using namespace std;

struct Node
{
	int distance,fuel;  //定义一个优先队列
	friend bool operator<(Node a, Node b)
	{     //从小到大排序采用“>”号；如果要从大到小排序，则采用“<”号
		return a.fuel < b.fuel;       //从大到小排序
	}
}node[10001];

bool cmp(const Node &a,const Node &b)
{
	return a.distance < b.distance;
}
void Solve(int n,int L,int P)
{
	int i=0,j=P,ans=0;
	priority_queue<Node>q;  //用优先队列来维护一个按照加油站油量从大到小来排序
	Node temp;
	while(node[i].distance<0)
		i++;
	while(i < n && j < L)
	{
		while(i < n && node[i].distance <= j)   //有油的时候能走多远就走多远
		{
			q.push(node[i]);
			++i;
		}
		if(q.empty())
			break;
		temp = q.top();
		q.pop();
		j += temp.fuel;   //如果走不到，就必须要加一次油，途中会遇到很多加油站，一定要选择油最多的那个加油站
		ans++;
	}
	if(j < L)
		printf("-1\n");
	else
		printf("%d\n",ans);
}
int main(void)
{
    int i,n,L,P;
	while (scanf("%d",&n)!=EOF)
	{
		sort(node,node+n,cmp);
		for (i = 0; i < n; ++i)
			scanf("%d %d",&node[i].distance,&node[i].fuel);
		scanf("%d %d",&L,&P);
		for (i = 0; i < n; ++i)
			node[i].distance = L - node[i].distance;   //当前位置跟加油站之间的距离
		sort(node,node+n,cmp);
		Solve(n,L,P);
	}
	return 0;
}

http://poj.org/problem?id=2465Adventures in Moving - Part IV

/*
题意:有一辆车要从起点0，到终点L处，中间有若干个加油站，给出车的油箱容量200，每行驶1km耗油1L。给出加油站的坐标，以及每个加油站的油价。一开始油箱里有100L的油，到达终点时必须还有100L的油，求最少花多少钱在加油上
分析:DP。F[i,j]表示到达第i个加油站有油j升的最优值，到达一个加油站时,枚举加多少油就好了
*/
//汽车油箱为200，初始油箱内只有100，当到达目的地时，油箱内的油>=100

#include<iostream>
#include<cstdio>
using namespace std;
#include<memory.h>

#define INF 1061109567

inline int min(int a,int b)
{
	if(a<b)
		return a;
	else
		return b;
}

int dist[103],price[103];
int dp[103][203];  //dp[i][j]表示到达地i个加油站，还剩下j油量的最少代价

int main(void)
{
	int i,j,k,n,len,d,p,m,ans;
	scanf("%d",&len);
	n=0;
	while(scanf("%d %d",&d,&p) != EOF)
	{
		dist[n] = d;
		price[n++] = p;
	}
	dist[n] = len;
	price[n++] = INF;
	memset(dp,0x3f3f3f,sizeof(dp));

	dp[0][100-dist[0]] = 0;
	for(i = 0 ; i < n - 1 ; ++i)
	{
		for(j = 0 ; j <= 200 ; ++j)
		{
			if(dp[i][j] == INF)
				continue;
			for(k = 0 ; k+j <= 200 ; ++k)
			{
				m = k + j - (dist[i+1] - dist[i]);
				if(m < 0 )
					continue;
				dp[i+1][m] = min(dp[i+1][m],dp[i][j] + k*price[i] );
			}
		}
	}
	ans = INF;
	for(i = 100 ; i <= 200 ; ++i)
	{
		if(dp[n-1][i] < ans)
			ans = dp[n-1][i];
	}
	if(ans == INF)
		puts("Impossible");
	else
		printf("%d\n",ans);
	return 0;
}

http://pat.zju.edu.cn/contests/pat-practise/1033 To Fill or Not to Fill

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
  
struct Node
{
	double price;
	double length;
}node[502];
bool cmp(const Node &a,const Node &b)
{
	return a.length<b.length;
}
double  capacity , dist ;
int unit_gas ,n;

int main(void)
{
    int i,j,m,index;
	double sum,len,cur_capacity,min_price;
	bool flag;
	while (scanf("%lf %lf %d %d",&capacity,&dist,&unit_gas,&n)!=EOF)
	{
		len = capacity*unit_gas;
		for (i = 0; i < n; ++i)
			scanf("%lf %lf",&node[i].price,&node[i].length);
		sort(node,node+n,cmp);
		node[n].price = 0;
		node[n].length = dist;

		if(node[0].length>0)
		{
			printf("The maximum travel distance = 0.00\n");
			continue;
		}
		else
		{
			flag = false;
			cur_capacity = 0;
			sum = 0;
			for( i = 0 ; i < n ;)
			{
				if(node[i+1].length - node[i].length > len)    //某两个油站之间的距离大于汽车油箱装满油量的最大行程
				{
					flag = true;
					printf("The maximum travel distance = %.2lf\n",node[i].length + len);
					break;
				}
				else
				{
					index = i;
					min_price = node[i].price;
					//找出当前油箱里的油能到达的所有加油站里，油价最便宜的那个
					for(j = i + 1 ; node[j].length - node[i].length <= cur_capacity*unit_gas && j <= n ; j++)
					{
						if(node[j].price < min_price)
						{
							min_price = node[j].price;
							index = j;
						}
					}
					if(index != i)
					{
						cur_capacity -= (node[index].length - node[i].length)/unit_gas;
						i = index;
						continue;
					}
					//若找不到，找出最近的一个能到达的比当前油价便宜的站，加一些油，跑到那个站
					index = i;
					for(j = i + 1 ; node[j].length - node[i].length <= len && j <= n ; j++)
					{
						if(node[j].price < node[i].price)
						{
							index = j;
							break;
						}
					}
					if(index != i)
					{
						sum += ((node[index].length - node[i].length)/unit_gas - cur_capacity)*node[i].price;
						cur_capacity = 0;
						i = index;
						continue;
					}
					//找不到比当前油站的价格还便宜的油站的时候
					//在当前油站需要加满油，跑到能跑到的所有站里油价最小的那个油站
					index = i;
					min_price = 1e18;
					for(j = i + 1 ; node[j].length - node[i].length <= len && j <= n ; j++)
					{
						if(node[j].price < min_price)
						{
							min_price = node[j].price;
							index = j;
						}
					}
					sum += (capacity-cur_capacity)*node[i].price;
					cur_capacity = capacity - (node[index].length - node[i].length )/unit_gas;
					i = index;
				}//else
			}//for
		}//else
		if(!flag)
			printf("%.2lf\n",sum);
	}
	return 0;
}

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