HDU—6024（Building Shops）

Building Shops

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1404    Accepted Submission(s): 525

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 thesen 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

题意描述：杭州电子科技大学有很多空的教室被考虑建成糖果屋，现在有n间特定教室可能被建为糖果屋，但是建造糖果屋需要一定量的花费，对于这n见教室，最左边的第一个教室肯定会被建成糖果屋，然后其右边的教室m如果选择修建糖果屋，则需要承担建糖果屋的费用，如果选择不建糖果屋，则相应花费为它到最左边糖果屋的距离，现给出特定教室的坐标，求总花费最小应该怎样修建糖果屋。

解题分析：首先，修建教室时会对其后的教室产生后效性，即当前修建的教室会对后面的教室修建产生影响，故我们采用DP的思路处理该问题，创建dp[100005][2]数组存储教室信息，dp[ i ][0]表示第i间教室不修建糖果屋的总费用，dp[ i ][1]表示第 i 间教室修建糖果屋的总费用。现在推导递推式，对于第 i 间教室，假定花费为 t ，则有     

dp[ i ][1] = min( dp[ i-1][0] , dp[ i -1][1] ) + t ，对于dp[ i ][0]，假定其左边最近糖果屋为第 j 间，距离为distant，则有

dp[ i ][0] = dp[ j ][1] + distant，所以，我们现在就需要找到这个第 j 间糖果屋，又糖果屋性质：dp[ j ][0] >= dp[j ][1]，

所以可以通过遍历找到 j , 相应遍历方法如下(site 为教室地点)：

for(j=i-1;j>=0;j--){dis = dis + (i-j)*(room[j+1].site - room[j].site);dp[i][0] = min( dp[i][0] ,(dp[j][1] + dis)); }

遍历代码解释: 从 i 的前一个教室开始遍历即 j = i - 1，如果 j 为糖果屋，则dp[ j ][0] >= dp[ j ][1]，进一步可知：

dp[i][0] 最终等于dp[ j ][1] + dis，如果 j 不是糖果屋，则继续遍历，其中距离的计算是从找到的糖果屋 j 到 i 之间的所有未修糖果屋距离之和。

AC code :

#include<stdio.h>#include<iostream>#include<algorithm>const int INF = 1234567890;using namespace std;struct cla{long long  site;long long  fare;};int cmp(cla a , cla b){return(a.site  < b.site );}int main (void){int count;while(scanf("%d",&count) != EOF){int i,j;long long ans = 0;//必用long long ;long long dp[5000][2];struct cla room[5000];for(i=0;i<count;i++){scanf("%lld%lld",&room[i].site , &room[i].fare);}sort(room,room+count,cmp); dp[0][0] = INF;dp[0][1] = room[0].fare; for(i=1;i<count;i++){long long dis = 0;dp[i][0] = INF;dp[i][1] = min(dp[i-1][0] ,dp[i-1][1]) + room[i].fare ;for(j=i-1;j>=0;j--){ dis = dis + (i-j)*(room[j+1].site - room[j].site); dp[i][0] = min( dp[i][0] ,(dp[j][1] + dis)); }}ans = min(dp[count-1][0],dp[count-1][1]);printf("%lld\n",ans);}}