HDU-5380 Travel with candy(贪心+单调队列)

Travel with candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 396    Accepted Submission(s): 194

Problem Description

There are n+1 cities on a line. They are labeled from city 0 to city n. Mph has to start his travel from city 0, passing city 1,2,3...n-1 in order and finally arrive city n. The distance between city i and city 0 is ai . Mph loves candies and when he travels one unit of distance, he should eat one unit of candy. Luckily, there are candy shops in the city and there are infinite candies in these shops. The price of buying and selling candies in city i is buyi and selli per unit respectively. Mph can carry at most C unit of candies.
Now, Mph want you to calculate the minimum cost in his travel plan.

Input

There are multiple test cases.
The first line has a number T, representing the number of test cases.
For each test :
The first line contains two numbers N and C (N≤2×105,C≤106)
The second line contains N numbers a1,a2,...,an . It is guaranteed that ai>ai−1 for each 1<i<=N .
Next N+1 line : the i-th line contains two numbers buyi−1 and selli−1 respectively. (selli−1≤buyi−1≤106 )

The sum of N in each test is less than 3×105

Output

Each test case outputs a single number representing your answer.(Note: the answer can be a negative number)

Sample Input

1 4 9 6 7 13 18 10 7 8 4 3 2 5 4 5 4

Sample Output

105

Author

SXYZ

Source

​​2015 Multi-University Training Contest 8 ​​

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思路：

1、在路上消耗的糖果一定是尽量最便宜的。

2、其它的部分我们可以带着准备卖。

那么每次离开一个点的时候，我们都把口袋补充满。

那么每次到达一个点的时候，我们口袋里都会剩下路途消耗之后的糖果。

此时：

1、口袋里那些购买价格高于当前点购买价格的糖果，我们可以当它们没有被买过，直接以买入价卖出就好。

2、口袋里那些购买价格低于当前点卖出价格的糖果，我们可以当它们有3种用途，第一种是后面被优先消耗了，第二种是在价格更高的点卖出了，第三种是到了最后剩下了。前两种可以忽视掉当前点，第三种则相当于这些糖果在这个点卖出了。那么我们就可以看做这些糖果在这个点就被卖出了，那么它们的买入价就可以看做这个点的卖出价。(意思就是，如果在这个点的后面还存在比当前点更优的情况来卖掉，没关系，这里先把卖掉，在用原价把买回来，相当于没有卖！)

或者换句话说，我买入了一个糖果，它的价值就是买入价，如果我带着它到了一个卖出价更高的地方，那么它的价值就成了卖出价。在路上我只消耗当前价值最低的糖果，如果一个糖果到了最后我都没有吃的话，那么就意味着我可以在它价值最高的地方把它卖掉。

贪心的绝世好题！思路忒难！虽然标题叫单调队列然而我并没有发现跟单调队列有关的东西 _(:зゝ∠)_ laj被几个long long 的强制转换卡的wa了还几次 _(:зゝ∠)_

1 #include "bits/stdc++.h"
 2 using namespace std;
 3 typedef long long LL;
 4 const int MAX=2e5+5;
 5 int T,n,m;
 6 int a[MAX],b[MAX],c[MAX];
 7 struct Node{int val,num;}q[MAX];
 8 int low,high;
 9 int main(){
10     freopen ("candy.in","r",stdin);
11     freopen ("candy.out","w",stdout);
12     int i,j;
13     scanf("%d",&T);
14     while (T--){
15         scanf("%d%d",&n,&m);n++;
16         a[0]=a[1]=0;
17         for (i=2;i<=n;i++) scanf("%d",a+i);
18         for (i=1;i<=n;i++) scanf("%d%d",b+i,c+i);
19         LL ans=0;
20         int dis,sum=0;
21         low=1,high=0;
22         for (i=1;i<=n;i++){
23             dis=a[i]-a[i-1];sum-=dis;
24             while (dis!=0){
25                 if (q[low].num<=dis){
26                     dis-=q[low].num;
27                     low++;
28                 }
29                 else{
30                     q[low].num-=dis;
31                     dis=0;
32                 }
33             }
34             for (j=low;j<=high;j++)
35                 q[j].val=max(q[j].val,c[i]);
36             while (low<=high && q[high].val>b[i]){
37                 ans-=(LL)q[high].val*(LL)q[high].num;
38                 sum-=q[high].num;
39                 high--;
40             }
41             if (sum<m){
42                 high++;
43                 q[high].val=b[i];q[high].num=m-sum;
44                 ans+=(LL)b[i]*(LL)(m-sum);
45                 sum=m;
46             }
47         }
48         while (low<=high) ans-=(LL)q[low].val*(LL)q[low].num,low++;
49         printf("%lld\n",ans);
50     }
51     return 0;
52