ZOJ-3469 Food Delivery

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 Bi 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 Nlines 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

题意：在一条x轴上有一些点，每个点会随着时间的增加点权不断增加，你每次要取一个点。问最后能获得的总点权最小是多少。

嗯，这个就是区间dp感觉比较复杂一些的？看大佬貌似只是基础题。。。。。

是我还是太天真了吧。

状态上必须加上左右端点。然后你每次可以从左边走或者从右边走。。。实际上是4种情况。

注意花费，这里的影响是当前的点增加的价值还有没取的点的一个价值。实际上为什么用dp就是因为这个影响。

看代码应该会更好理解一些。。。

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <time.h>

using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 1000+7;
int n,v,x;
int sum[MAXN];
int dp[MAXN][MAXN][2];
int get_cost(int l,int r)
{
    if(l > r)return 0;
    return sum[r] - sum[l-1];
}
struct node
{
    int x,v;
    bool operator < (const node &a)const
    {
        return x < a.x;
    }
}p[MAXN];

int main()
{
    while(~scanf("%d%d%d",&n,&v,&x))
    {
        sum[0] = 0;
        memset(dp,inf,sizeof dp);
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d%d",&p[i].x,&p[i].v);
        }
        p[++n].x = x;
        p[n].v = 0;
        sort(p+1,p+1+n);
        int pos = 1;
        for(int i = 1; i <= n; ++i)
        {
            sum[i] = sum[i-1] + p[i].v;
            if(p[i].x == x)pos = i;
        }
        
        dp[pos][pos][0] = dp[pos][pos][1] = 0;
        for(int i = pos; i >= 1; --i)
            for(int j = pos; j <= n; ++j)
        {
            int cost = get_cost(1,i-1) + get_cost(j+1,n);
            dp[i][j][0] = min(dp[i][j][0],dp[i+1][j][0] + (p[i+1].x - p[i].x)*(p[i].v + cost));
            dp[i][j][0] = min(dp[i][j][0],dp[i+1][j][1] + (p[j].x - p[i].x)*(p[i].v + cost));
            dp[i][j][1] = min(dp[i][j][1],dp[i][j-1][0] + (p[j].x - p[i].x)*(p[j].v + cost));
            dp[i][j][1] = min(dp[i][j][1],dp[i][j-1][1] + (p[j].x - p[j-1].x)*(p[j].v + cost));
        }
        printf("%d\n",min(dp[1][n][0],dp[1][n][1])*v);
    }
    return 0;
}
           

然后感觉数位dp题型比较复杂啊，虽然比线性dp要好一些吧，还是先做多一些题看看。