链接

题解链接：http://www.cygmasot.com/index.php/2015/08/16/hdu_5380

题意：

n C

一条数轴上有n+1个加油站，起点在0，终点在n。车的油箱容量为C

下面n个数字表示每个加油站距离起点的距离。

下面n+1行表示每个加油站买进和卖出一单位油的价格。油可以买也可以卖。

问开到终点的最小花费。

思路:

把油箱保持装满，然后维护一个价格单调递增的队列。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e9;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x < 0) { putchar('-'); x = -x; }
	if (x > 9) pt(x / 10);
	putchar(x % 10 + '0');
}
using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
typedef pair<int, int> pii;
ll n, c;
ll in[N], out[N], dis[N];
ll work() {
	ll ans = c*in[0];
	deque<ll> In, o;
	In.push_back(in[0]); o.push_back(c);
	for (int i = 1; i <= n; i++)
	{
		ll tmp = dis[i];
		while (tmp) {
			ll LEF = o.front();
			ll mi = min(tmp, LEF);
			LEF -= mi;
			tmp -= mi;
			o.pop_front();
			if(LEF)
				o.push_front(LEF);
			else
				In.pop_front();
		}
		ll tot = dis[i]; tmp = 0;
		while (!In.empty()) {
			if (In.front() <= out[i])
			{
				tmp += o.front();
				ans -= out[i]*o.front();
				o.pop_front();
				In.pop_front();
			}
			else break;
		}
		if (tmp) {
			ans += out[i] * tmp;
			o.push_front(tmp);
			In.push_front(out[i]);
		}
		while (!In.empty()) {
			if (In.back() >= in[i])
			{
				tot += o.back();
				ans -= In.back()*o.back();
				o.pop_back();
				In.pop_back();
			}
			else break;
		}
		o.push_back(tot);
		In.push_back(in[i]);
		ans += in[i] * tot;
	}
	while (!In.empty()) {
		ans -= o.front() * In.front();
		In.pop_front(); o.pop_front();
	}
	return ans;
}
int main() {
	int T; rd(T);
	while (T--) {
		rd(n); rd(c);
		for (int i = 1; i <= n; i++)rd(dis[i]);
		for (int i = n; i > 1; i--)dis[i] -= dis[i - 1];
		for (int i = 0; i <= n; i++)rd(in[i]), rd(out[i]);
		pt(work()); puts("");
	}
	return 0;
}
/*
1
3 9
2 4 6
8 2
4 3
6 3
9 6

1
4 9
4 9 12 18
5 1
7 6
3 2
4 2
8 6

1
4 5
2 4 8 10
2 1
2 1
9 3
9 8
7 2

1
9 4
2 4 5 7 8 9 11 14 15
9 8
10 5
8 2
4 3
2 1
7 3
9 6
10 8
5 3
8 5

*/

Travel with candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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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