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; }