Food Delivery ZOJ

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. Thei^th person's coordinate is X_i meters. And in the street there is a take-out restaurant which has coordinatesX 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 theN 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 byDispleasure Index. At the beginning, the Displeasure Index for each person is 0. When waiting for the food, thei^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'sDispleasure Index as low as possible in order to maximize your income. Your task is to find the minimal sum ofDispleasure Index.

Input

The input contains multiple test cases, separated with a blank line. Each case is started with three integersN ( 1 <= N <= 1000 ), V ( V > 0), X (X >= 0 ), then N lines followed. Each line contains two integersX_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

55

恩　没错，这道题我又看题解了，不过我感觉我已经有所突破。

这道题就是分析题。

首先如果在一个地方有个餐厅要去送餐，那么送完了一定存在两个状态，不是在区间的左边就是在区间的右边

这样的话我们已餐厅的位置为首要位置向两边扩充即可

对于dp[i][j][0]他的状态只能通过dp[i+1][j][0]转移而不能通过dp[i+1][j][1]转移而来，因为如果从右边来的话，再到左边一定遍历了左边的所求的区间，这样的话一定不是最优的，所有每一种状态对应的转移方程是一定的。

所以ｄｐ问题主要靠分析问题的本质

#include<bits/stdc++.h>using namespace std;int dp[1086][1086][2];int sum[1086];struct aa{    int pos;    int val;} sv[1086];bool cmp(const aa &a,const aa &b){    return a.pos<b.pos;}int main(){    int n,v,p,pos;    while(scanf("%d%d%d",&n,&v,&p)!=EOF)    {        for(int i=1; i<=n; i++)            scanf("%d%d",&sv[i].pos,&sv[i].val);        sv[++n].pos=p;        sv[n].val=0;        memset(dp,0,sizeof(dp));        sort(sv+1,sv+1+n,cmp);        for(int i=1; i<=n; i++)        {            if(sv[i].pos==p)            {                pos=i;                break;            }        }        memset(dp,0x3f3f3f3f,sizeof(dp));        sum[0]=0;        for(int i=1; i<=n; i++)        {            sum[i]=sum[i-1]+sv[i].val;        }        dp[pos][pos][0]=dp[pos][pos][1]=0;        for(int i = pos; i >= 1; i--)        {            for(int j = pos; j <= n; j++)            {                int delay = sum[i-1]+sum[n]-sum[j];                if(i == j)                    continue;                dp[i][j][0] = min(dp[i][j][0], dp[i+1][j][0]+(sv[i+1].pos-sv[i].pos)*(delay+sv[i].val));                dp[i][j][0] = min(dp[i][j][0], dp[i+1][j][1]+(sv[j].pos-sv[i].pos)*(delay+sv[i].val));                dp[i][j][1] = min(dp[i][j][1], dp[i][j-1][0]+(sv[j].pos-sv[i].pos)*(delay+sv[j].val));                dp[i][j][1] = min(dp[i][j][1], dp[i][j-1][1]+(sv[j].pos-sv[j-1].pos)*(delay+sv[j].val));            }        }        printf("%d\n", min(dp[1][n][0], dp[1][n][1])*v);    }}