汽车加油行驶问题

http://poj.org/problem?id=2431   Expedition

/*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");elseprintf("%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=2465    Adventures 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 1061109567inline int min(int a,int b){if(a<b)return a;elsereturn 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");elseprintf("%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}//elseif(!flag)printf("%.2lf\n",sum);}return 0;}