UVA 1169(dp + 单调队列)

Problem C - Robotruck

Background

This problem is about a robotic truck that distributes mail packages to several locations in a factory. The robot sits at the end of a conveyer at the mail office and waits for packages to be loaded into its cargo area. The robot has a maximum load capacity, which means that it may have to perform several round trips to complete its task. Provided that the maximum capacity is not exceeded, the robot can stop the conveyer at any time and start a round trip distributing the already collected packages. The packages must be delivered in the incoming order.

The distance of a round trip is computed in a grid by measuring the number of robot moves from the mail office, at location (0,0), to the location of delivery of the first package, the number of moves between package delivery locations, until the last package, and then the number of moves from the last location back to the mail office. The robot moves a cell at a time either horizontally or vertically in the factory plant grid. For example, consider four packages, to be delivered at the locations (1,2), (1,0), (3,1), and (3,1). By dividing these packages into two round trips of two packages each, the number of moves in the first trip is 3+2+1=6, and 4+0+4=8 in the second trip. Notice that the two last packages are delivered at the same location and thus the number of moves between them is 0.

Problem

Given a sequence of packages, compute the minimum distance the robot must travel to deliver all packages.

Input

Input consists of multiple test cases the first line of the input contains the number of test cases. There is a blank line before each dataset. The input for each dataset consists of a line containing one positive integer C, not greater then 100, indicating the maximum capacity of the robot, a line containing one positive integer N, not greater than 100,000, which is the number of packages to be loaded from the conveyer. Next, there are Nlines containing, for each package, two non-negative integers to indicate its delivery location in the grid, and a positive integer to indicate its weight. The weight of the packages is always smaller than the robot’s maximum load capacity. The order of the input is the order of appearance in the conveyer.

Output

One line containing one integer representing the minimum number of moves the robot must travel to deliver all the packages. Print a blank line between datasets.

Sample Input

1

10

4

1 2 3

1 0 3

3 1 4

3 1 4

Sample Output

14

题意：有一个机器人，给定w和n，w代表机器人的最大负重，n代表坐标系上有n个垃圾，接下去n行代表垃圾的位置和重量，一开始机器人在原点，机器人只能走横竖，机器人可以拿多个垃圾，只要不超过最大负重，然后放回垃圾桶，垃圾桶在原点，问机器人最少需要走的距离。

思路：n非常的大，没什么思路，看了刘汝佳的代码，原来是加入了单调队列去维护。dp[i]表示前i个垃圾需要的最少距离，然后先预处理出sumw[i]，前i个垃圾的重量之和，sumd[i]，路过前i个垃圾的距离之和，d[i]，i到原点的哈曼顿距离那么对于j到i的垃圾，sumw = sumw[i] - sumw[j], sumd = sumd[i] - sumd[j]。由于对于dp[i]而言，最后机器人肯定是回到原点了，所以这样一来可以推出状态转移方程为dp[i] = dp[j] + d[j + 1] + sumd + d[i] (if (sumw <= w)。如果单单这样看的话，需要两重for循环，复杂度是o(n^2)超时。那么把方程转换一下变成: dp[i] = dp[j] + d[j + 1] - sumd[j + 1] + sumd[i] + d[i]; 设dp[j] + d[j + 1] - sumd[j + 1] = func(j);那么对于每一个状态dp[i] = func(j) + sumd[i] + d[i]; 要保证最小，只要维护func(j)的最小值即可。这步就可以使用一个单调队列来维护，使得复杂度降为o(n)。很巧妙的方法。

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <queue>
using namespace std;
const int N = 100005;

int t, w, n, sumw[N], sumd[N], dp[N];
struct Point {
	int x, y, w, d;
	Point() {x = 0; y = 0; w = 0; d = 0;}
} p[N], zero;;

int dis(Point a, Point b) {
	return abs(a.x - b.x) + abs(a.y - b.y);
}

int func(int j) {
	return dp[j] - sumd[j + 1] + p[j + 1].d;
}

int solve() {
	deque<int> Q;
	Q.push_front(0);
	for (int i = 1; i <= n; i++) {
		while (!Q.empty() && sumw[i] - sumw[Q.front()] > w) {Q.pop_front();}
		dp[i] = func(Q.front()) + sumd[i] + p[i].d;
		while (!Q.empty() && func(i) <= func(Q.back())) {Q.pop_back();}
		Q.push_back(i);
	}
	return dp[n];
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &w, &n);
		for (int i = 1; i <= n; i++) {
			scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].w);
			sumw[i] = sumw[i - 1] + p[i].w;
			sumd[i] = sumd[i - 1] + dis(p[i], p[i - 1]);
			p[i].d = dis(p[i], zero);
		}
		printf("%d\n", solve());
		if (t) printf("\n");
	}
	return 0;
}