hdu6024（dp）
Building Shops
HDU’s nn 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 nn classrooms.
The total cost consists of two parts. Building a candy shop at classroom ii would have some cost cici. For every classroom PP without any candy shop, then the distance between PP and the rightmost classroom with a candy shop on PP’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)n(1≤n≤3000), denoting the number of the classrooms.
In the following nn lines, each line contains two integers xi,ci(−109≤xi,ci≤109)xi,ci(−109≤xi,ci≤109), denoting the coordinate of the ii-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][1]表示第i个教室开超市，dp[i][1]=min(dp[i-1][0],dp[i-1][1])+cost[i]
dp[i][0]表示第i个教室不开超市，这时遍历i前的所有教室，计算距离，更新dp[i][0]的值
别忘了先按坐标排个序
#include <bits/stdc++.h>
using namespace std;
const long long inf=1e12;
struct hh
{
long long x,c;
bool operator < (hh&a)
{
return this->x<a.x;
}
hh():x(0),c(0){
}
}s[3005];
long long dp[3005][2];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
{
scanf("%lld%lld",&s[i].x,&s[i].c);
}
sort(s,s+n);
memset(dp,0,sizeof(dp));
dp[0][0]=inf;
dp[0][1]=s[0].c;
for(int i=1;i<n;i++)
{
dp[i][1]=min(dp[i-1][0],dp[i-1][1])+s[i].c;
dp[i][0]=inf;
long long t=0;
for(int j=i-1;j>=0;j--)
{
t+=(i-j)*(s[j+1].x-s[j].x);
dp[i][0]=min(dp[i][0],dp[j][1]+t);
}
}
cout<<min(dp[n-1][0],dp[n-1][1])<<'\n';
}
return 0;
}
