Hdu6024 dp(类01背包)

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

31 22 33 441 73 15 106 1

 

Sample Output

511

#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>using namespace std;typedef long long ll;const ll INF=1e18;struct node{int c;int d;}a[3030];ll dis[3010][3010];ll dp[3010][5];bool cmp(node n1,node n2){return n1.d<n2.d;}int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++)scanf("%d%d",&a[i].d,&a[i].c);sort(a,a+n,cmp);int i,j;for(i=0;i<n;i++){dis[i][i]=0;for(j=i+1;j<n;j++){dis[i][j]=dis[i][j-1]+a[j].d-a[i].d;}}for(int i=0;i<n;i++){dp[i][0]=dp[i][1]=INF;} dp[0][1]=a[0].c;for(i=1;i<n;i++){dp[i][1]=min(dp[i-1][0],dp[i-1][1])+a[i].c;for(j=i-1;j>=0;j--){dp[i][0]=min(dp[i][0],dp[j][1]+dis[j][i]);}}printf("%lld\n",min(dp[n-1][0],dp[n-1][1]));  }}