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
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
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！！！！）
1 /* 2 qwerta 3 P3080 [USACO13MAR]牛跑The Cow Run 4 Accepted 5 100 6 代码 C++，0.81KB 7 提交时间 2018-09-22 18:59:00 8 耗时/内存 9 231ms, 16760KB 10 */ 11 #include<cmath> 12 #include<cstdio> 13 #include<cstring> 14 #include<iostream> 15 #include<algorithm> 16 using namespace std; 17 int dis[1007]; 18 long long f[1007][1007][2]; 19 int main() 20 { 21 //freopen("a.in","r",stdin); 22 int n; 23 scanf("%d",&n); 24 for(int i=1;i<=n;++i) 25 scanf("%d",&dis[i]); 26 n++; 27 sort(dis+1,dis+n+1); 28 memset(f,127,sizeof(f)); 29 int x=0; 30 for(int i=1;i<=n;++i) 31 if(dis[i]==0)x=i; 32 f[x][x][0]=f[x][x][1]=0; 33 for(int len=2;len<=n;++len) 34 for(int l=1,r=len;r<=n;++l,++r) 35 { 36 f[l][r][0]=min(f[l+1][r][0]+(dis[l+1]-dis[l])*(n-len+1), 37 f[l+1][r][1]+(dis[r]-dis[l])*(n-len+1)); 38 f[l][r][1]=min(f[l][r-1][0]+(dis[r]-dis[l])*(n-len+1), 39 f[l][r-1][1]+(dis[r]-dis[r-1])*(n-len+1)); 40 } 41 cout<<min(f[1][n][0],f[1][n][1]); 42 return 0; 43 }