ZOJ 3469 - Food Delivery （区间DP）

外卖配送最优路径算法

最新推荐文章于 2020-08-10 20:38:05 发布

原创最新推荐文章于 2020-08-10 20:38:05 发布 · 270 阅读

0 ·

CC 4.0 BY-SA版权

ZOJ 同时被 2 个专栏收录

21 篇文章

订阅专栏

区间DP

9 篇文章

订阅专栏

本文介绍了一种解决外卖配送问题的算法，通过动态规划的方法找到使顾客不满情绪总和最小的配送路径。考虑了顾客的位置、不满情绪增长速度等因素。

Food Delivery

Time Limit: 2 Seconds Memory Limit: 65536 KB

When we are focusing on solving problems, we usually prefer to stay in front of computers rather than go out for lunch. At this time, we may call for food delivery.

Suppose there are N people living in a straight street that is just lies on an X-coordinate axis. The i^th person's coordinate is X_i meters. And in the street there is a take-out restaurant which has coordinates X meters. One day at lunchtime, each person takes an order from the restaurant at the same time. As a worker in the restaurant, you need to start from the restaurant, send food to the N people, and then come back to the restaurant. Your speed is V^-1 meters per minute.

You know that the N people have different personal characters; therefore they have different feeling on the time their food arrives. Their feelings are measured by Displeasure Index. At the beginning, the Displeasure Index for each person is 0. When waiting for the food, the i^th person will gain B_i Displeasure Index per minute.

If one's Displeasure Index goes too high, he will not buy your food any more. So you need to keep the sum of all people's Displeasure Index as low as possible in order to maximize your income. Your task is to find the minimal sum of Displeasure Index.

Input

The input contains multiple test cases, separated with a blank line. Each case is started with three integers N ( 1 <= N <= 1000 ), V ( V > 0), X ( X >= 0 ), then N lines followed. Each line contains two integers X_i ( X_i >= 0 ), B_i ( B_i >= 0), which are described above.

You can safely assume that all numbers in the input and output will be less than 2³¹ - 1.

Please process to the end-of-file.

Output

For each test case please output a single number, which is the minimal sum of Displeasure Index. One test case per line.

Sample Input

5 1 0
1 1
2 2
3 3
4 4
5 5

Sample Output

Author: LI, Cheng
Contest: ZOJ Monthly, February 2011

题意：

一维坐标轴上，有n个人点了外卖，店家在x点上。每个人在xi等待，等待一分钟会长bi个愤怒值。

问你送完后愤怒值最少是多少。

POINT：

要知道DP的区间肯定要包含店家这个点。不然没有意义。

DP[x][y][0]代表送完x到y这个区间后店家在x

DP[x][y][1]代表送完x到y这个区间后店家在y

愤怒值怎么算：每次把没送到的人的愤怒值全部加上。即区间外的人。

DP做：

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define  LL long long
const LL maxn = 1e3+55;
const LL mod = 1e9+7;
const LL inf = 1e18;
LL to = 0;
struct node
{
	LL x,b;

	bool friend operator < (node a,node b)
	{
		return a.x<b.x;
	}

}a[maxn];

LL dp[1111][1111][2];
LL n,v,x;
LL sum[1111];


LL Fsum(LL l,LL r)
{
	return sum[n]-(sum[r]-sum[l-1]);
}

LL dis(LL l,LL r)
{
	return abs(a[r].x-a[l].x);
}

int main()
{

	while(~scanf("%lld %lld %lld",&n,&v,&x)){
		for(LL i=1;i<=n;i++){
			scanf("%lld %lld",&a[i].x,&a[i].b);
		}
		a[++n].x=x;
		a[n].b=0;
		sort(a+1,a+1+n);
		for(int i=1;i<=n;i++){
			sum[i]=sum[i-1]+a[i].b;
			for(int j=1;j<=n;j++)
				dp[i][j][0]=dp[i][j][1]=inf;
		}
		for(LL i=1;i<=n;i++){
			if(x==a[i].x){
				to=i;
				break;
			}
		}
		dp[to][to][0]=dp[to][to][1]=0;
		for(LL l=to;l>=1;l--){
			for(LL r=to;r<=n;r++){
				if(l==r) continue;
				dp[l][r][0]=min(dp[l][r][0],dp[l+1][r][0]+dis(l,l+1)*v*Fsum(l+1,r));
				dp[l][r][0]=min(dp[l][r][0],dp[l+1][r][1]+dis(l,r)*v*Fsum(l+1,r));
				dp[l][r][1]=min(dp[l][r][1],dp[l][r-1][0]+dis(l,r)*v*Fsum(l,r-1));
				dp[l][r][1]=min(dp[l][r][1],dp[l][r-1][1]+dis(r-1,r)*v*Fsum(l,r-1));


			}
		}

		printf("%lld\n",min(dp[1][n][0],dp[1][n][1]));
	}



}

DFS形式：

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define  LL long long
const LL maxn = 1e3+55;
const LL mod = 1e9+7;
const LL inf = 1e18;
LL to = 0;
struct node
{
	LL x,b;

	bool friend operator < (node a,node b)
	{
		return a.x<b.x;
	}

}a[maxn];

LL dp[1111][1111][2];
LL n,v,x;
LL sum[1111];
int vis[1111][1111];

void dfs(LL l,LL r)
{
	if(vis[l][r]) return;
	if(!(l<=to&&r>=to)) return;
	if(l==r&&l==to){
		dp[l][r][0]=dp[l][r][1]=0;
		return;
	}
	dfs(l+1,r);
	dfs(l,r-1);
	dp[l][r][0]=min(dp[l][r][0],dp[l+1][r][0]+(a[l+1].x-a[l].x)*v*(sum[n]-(sum[r]-sum[l])));
	dp[l][r][0]=min(dp[l][r][0],dp[l+1][r][1]+(a[r].x-a[l].x)*v*(sum[n]-(sum[r]-sum[l])));
	dp[l][r][1]=min(dp[l][r][1],dp[l][r-1][0]+(a[r].x-a[l].x)*v*(sum[n]-(sum[r-1]-sum[l-1])));
	dp[l][r][1]=min(dp[l][r][1],dp[l][r-1][1]+(a[r].x-a[r-1].x)*v*(sum[n]-(sum[r-1]-sum[l-1])));
	vis[l][r]=1;
}

int main()
{

	while(~scanf("%lld %lld %lld",&n,&v,&x)){
		for(LL i=1;i<=n;i++){
			scanf("%lld %lld",&a[i].x,&a[i].b);
		}
		a[++n].x=x;
		a[n].b=0;
		sort(a+1,a+1+n);
		for(int i=1;i<=n;i++){
			sum[i]=sum[i-1]+a[i].b;
			for(int j=1;j<=n;j++)
				vis[i][j]=0,dp[i][j][0]=dp[i][j][1]=inf;
		}
		for(LL i=1;i<=n;i++){
			if(x==a[i].x){
				to=i;
				break;
			}
		}
		dfs(1,n);
		printf("%lld\n",min(dp[1][n][0],dp[1][n][1]));
	}



}