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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

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

105

HDU 5380 Travel with candy 单调队列

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原文地址：http://blog.csdn.net/qq574857122/article/details/47702915