Building Shops
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 26 Accepted Submission(s): 16
Problem Description
HDU’s
n
classrooms are on a line ,which can be considered as a number line. Each classroom has a coordinate. Now Little Q wants to build several candy shops in these
n
classrooms.
The total cost consists of two parts. Building a candy shop at classroom
i
would have some cost
ci
. For every classroom
P
without any candy shop, then the distance between
P
and the rightmost classroom with a candy shop on
P
's left side would be included in the cost too. Obviously, if there is a classroom without any candy shop, there must be a candy shop on its left side.
Now Little Q wants to know how to build the candy shops with the minimal cost. Please write a program to help him.
Input
The input contains several test cases, no more than 10 test cases.
In each test case, the first line contains an integer
n(1≤n≤3000)
, denoting the number of the classrooms.
In the following
n
lines, each line contains two integers
xi,ci(−109≤xi,ci≤109)
, denoting the coordinate of the
i
-th classroom and the cost of building a candy shop in it.
There are no two classrooms having same coordinate.
Output
For each test case, print a single line containing an integer, denoting the minimal cost.
Sample Input
3
1 2
2 3
3 4
4
1 7
3 1
5 10
6 1
Sample Output
5
11
dp[i]表示建到i（且i建商店）的代价；
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b) memset(a,b,sizeof(a))
const int M=1e4+10;
const int inf=0x3f3f3f3f;
int i,j,k,n,m;
ll dp[M];
struct node
{
ll x,c;
}p[M];
bool cmp(node a,node b)
{
return a.x<b.x;
}
ll len[M];
int main()
{
int T;
while(~scanf("%d",&n)){
for(int i=1;i<=n;i++)
scanf("%lld%lld",&p[i].x,&p[i].c);
sort(p+1,p+n+1,cmp);
len[0]=0;
for(int i=1;i<=n;i++)
len[i]=len[i-1]+p[i].x-p[1].x;
ms(dp,inf);
dp[1]=p[1].c;
for(int i=2;i<=n;i++){
for(int j=1;j<i;j++){
if(j==i-1)dp[i]=min(dp[i],dp[j]+p[i].c);
else dp[i]=min(dp[i],dp[j]+p[i].c+len[i-1]-len[j]-(i-1-j)*(p[j].x-p[1].x));
}
}
ll ans=inf;
for(int i=1;i<=n;i++){
if(i==n)ans=min(ans,dp[n]);
else ans=min(ans,dp[i]+len[n]-len[i]-(n-i)*(p[i].x-p[1].x));
}
printf("%lld\n",ans);
}
return 0;
}
