「USACO13MAR」「LuoguP3080」牛跑The Cow Run （区间dp

题目描述

Farmer John has forgotten to repair a hole in the fence on his farm, and his N cows (1 <= N <= 1,000) have escaped and gone on a rampage! Each minute a cow is outside the fence, she causes one dollar worth of damage. FJ must visit each cow to install a halter that will calm the cow and stop the damage.

Fortunately, the cows are positioned at distinct locations along a straight line on a road outside the farm. FJ knows the location P_i of each cow i (-500,000 <= P_i <= 500,000, P_i != 0) relative to the gate (position 0) where FJ starts.

FJ moves at one unit of distance per minute and can install a halter instantly. Please determine the order that FJ should visit the cows so he can minimize the total cost of the damage; you should compute the minimum total damage cost in this case.

农夫约翰的牧场围栏上出现了一个洞，有N(1 <= N <= 1,000)只牛从这个洞逃出了牧场。这些出逃的奶牛很狂躁，他们在外面到处搞破坏，每分钟每头牛都会给约翰带来1美元的损失。约翰必须用缰绳套住所有的牛，以停止他们搞破坏。

幸运的是，奶牛们都在牧场外一条笔直的公路上，牧场的大门恰好位于公里的0点处。约翰知道每头牛距离牧场大门的距离P_i(-500,000 <= P_i <= 500,000, P_i != 0)

约翰从农场大门出发，每分钟移动一个单位距离，每到一头牛所在的地点，约翰就会给它套上缰绳，套缰绳不花时间。按怎样的顺序去给牛套缰绳才能使约翰损失的费用最少？

输入输出格式

输入格式：

* Line 1: The number of cows, N.

* Lines 2..N+1: Line i+1 contains the integer P_i.

输出格式：

* Line 1: The minimum total cost of the damage.

输入输出样例

输入样例#1：
复制

输出样例#1： 复制

说明

Four cows placed in positions: -2, -12, 3, and 7.

The optimal visit order is -2, 3, 7, -12. FJ arrives at position -2 in 2 minutes for a total of 2 dollars in damage for that cow.

He then travels to position 3 (distance: 5) where the cumulative damage is 2 + 5 = 7 dollars for that cow.

He spends 4 more minutes to get to 7 at a cost of 7 + 4 = 11 dollars for that cow.

Finally, he spends 19 minutes to go to -12 with a cost of 11 + 19 = 30 dollars.

The total damage is 2 + 7 + 11 + 30 = 50 dollars.

题解

不知道为什么能蓝。

这是一类被我们机房称为老王关灯的题鬼嘞只有我这么叫

老(wan)王(e)关(zhi)灯(yuan)

这道题比老王关灯还简单一丢丢，不用算功率啦，

单位时间消耗的dollar直接算未覆盖区间就好啦～

然后这种题开个long long开不出吃亏开不出上当

还能有效预防越界（QAQ！！！！）

 /*

     qwerta

     P3080 [USACO13MAR]牛跑The Cow Run

     Accepted

     100

     代码 C++，0.81KB

     提交时间 2018-09-22 18:59:00

     耗时/内存

     231ms, 16760KB

 */

 #include<cmath>

 #include<cstdio>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

 using namespace std;

 int dis[];

 long long f[][][];

 int main()

 {

     //freopen("a.in","r",stdin);

     int n;

     scanf("%d",&n);

     for(int i=;i<=n;++i)

     scanf("%d",&dis[i]);

     n++;

     sort(dis+,dis+n+);

     memset(f,,sizeof(f));

     int x=;

     for(int i=;i<=n;++i)

     if(dis[i]==)x=i;

     f[x][x][]=f[x][x][]=;

     for(int len=;len<=n;++len)

     for(int l=,r=len;r<=n;++l,++r)

     {

         f[l][r][]=min(f[l+][r][]+(dis[l+]-dis[l])*(n-len+),

                        f[l+][r][]+(dis[r]-dis[l])*(n-len+));

         f[l][r][]=min(f[l][r-][]+(dis[r]-dis[l])*(n-len+),

                        f[l][r-][]+(dis[r]-dis[r-])*(n-len+));

     }

     cout<<min(f[][n][],f[][n][]);

     return ;

 }