Description
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 ith person’s coordinate is Xi 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 ith person will gain BiDispleasure 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 Xi ( Xi >= 0 ), Bi ( Bi >= 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
55
1）题意：有一家快餐店送外卖，现在同时有n个家庭打进电话订购，送货员得以V-1的速度一家一家的运送，但是每一个家庭都有一个不开心的值，每分钟都会增加一倍，值达到一定程度，该家庭将不会再订购外卖了，现在为了以后有更多的家庭订购，要将外卖送到的情况下使得所有用户的不开心值总和达到最小。
（2）解法：区间dp。很明显，每多走一分钟，没送到的家庭的不开心值都会加倍。假设是这样的顺序123X456，从X出发先往左右中间靠近的送，再往两边送省时间。dp[i][j][0]表示从i到j用户送到最小不开心值，此时送货员停留在左边即i位置。dp[i][j][1]表示从i到j用户送到最小不开心值，此时送货员停留在右边即j位置状态有四种：
1）：dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][0]+(a[i+1].x-a[i].x)*(delay+a[i].v));
2）：dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][1]+(a[j].x-a[i].x)*(delay+a[i].v));
3）：dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][0]+(a[j].x-a[i].x)*(delay+a[j].v));
4）：dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][1]+(a[j].x-a[j-1].x)*(delay+a[j].v));
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
long long int dp[1005][1005][2];
int sum[1005];
struct Rac{
int x;
int bu;
}order[1005];
bool cmp(struct Rac a,struct Rac b)
{
return a.x<b.x;
}
int main()
{
int n,v,qi;
//freopen("a.txt","r",stdin);
while(scanf("%d %d %d",&n,&v,&qi)==3)
{
int i,j;
int zou;
memset(sum,0,sizeof(sum));
for(i=1;i<=n;i++) scanf("%d %d",&order[i].x,&order[i].bu);
n++;
order[n].x=qi;
order[n].bu=0;
sort(order+1,order+n+1,cmp);
for(i=1;i<=n;i++) sum[i]=sum[i-1]+order[i].bu;
for(i=1;i<=n;i++)
{
if(order[i].x==qi) zou=i;
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
dp[i][j][0]=dp[i][j][1]=1100000000;
}
}
dp[zou][zou][0]=dp[zou][zou][1]=0;
for(i=zou;i>=1;i--)
{
for(j=zou;j<=n;j++)
{
if(i==j) continue;
// sum[i]+sum[n]-sum[j]//剩下没有送的。。+fuck*t
dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][0]+(order[i+1].x-order[i].x)*(sum[i]+sum[n]-sum[j]));
dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][1]+(order[j].x-order[i].x)*(sum[i]+sum[n]-sum[j]));
dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][1]+(order[j].x-order[j-1].x)*(sum[i-1]+sum[n]-sum[j-1]));
dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][0]+(order[j].x-order[i].x)*(sum[i-1]+sum[n]-sum[j-1]));
}
}
printf("%d\n",min(dp[1][n][0],dp[1][n][1])*v);
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。