zoj 3469 Food Delivery（区间dp）

题目链接

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 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

越刷题越觉得自己菜的不行==

题目大意是说送外卖，每个客人都有一个不满意值，随时间增长而增长，问如何送能使所有人的不满意值最小。
首先我们可以想到外卖员从i处送到j处时，这条路径上的所有客人都一定会被送到。所以我们可以想到，外卖小哥的路线一定是以餐馆为中心来回送的（最好的情况下一遍就可以），那么送完ij这段路后，会有两种选择下一个送i-1或者j+1，我们可以发现接下来送哪个点最优还和另外一个变量有关，那就是送完ij后外卖员的位置，实际上位置只有两种情况，在i点或者j点，所以我们需要开一个三维的dp数组，而第三维只要记录两个状态即可。
然后我们以餐厅为中心，依次枚举每个区间。

#include<cstdio>#include<cstdlib>#include<cstring>#include <cmath>#include<iostream>#include<algorithm>#include <vector>#include <bitset>#include <stack>#define maxn 1010#define ll long long#define MEM(x,num) memset(x,num,sizeof(x))#define inf 0x3f3f3f3f#define mod 1000000007using namespace std;int dp[maxn][maxn][2];      //0代表送餐结束后在i处，1代表在j处int sum[maxn];int n,v,x;struct node{    int b,x;} a[maxn];bool cmp(node p,node q){    return p.x<q.x;}int main(){    while(cin>>n>>v>>x)    {        memset(dp,-1,sizeof(dp));        for(int i=1; i<=n; i++)            cin>>a[i].x>>a[i].b;        n++;        a[n].x=x;        a[n].b=0;        sort(a+1,a+1+n,cmp);        for(int i=1; i<=n; i++)            sum[i]=sum[i-1]+a[i].b;        memset(dp,0x3f,sizeof(dp));        int res;        for(int i=1; i<=n; i++)            if(a[i].x==x)            {                res=i;                break;            }        dp[res][res][0]=dp[res][res][1]=0;        for(int i=res; i>=0; i--)            for(int j=res; j<=n; j++)            {                if(i==j) continue;                dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][0]+(sum[n]-sum[j]+sum[i])*(a[i+1].x-a[i].x));                dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][1]+(sum[n]-sum[j]+sum[i])*(a[j].x-a[i].x));                dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][1]+(sum[n]-sum[j-1]+sum[i-1])*(a[j].x-a[j-1].x));                dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][0]+(sum[n]-sum[j-1]+sum[i-1])*(a[j].x-a[i].x));            }        cout<<v*min(dp[1][n][0],dp[1][n][1])<<endl;    }    return 0;}