题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6024
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 test cases.
In each test case, the first line contains an integer n(≤n≤), denoting the number of the classrooms.
In the following n lines, each line contains two integers xi,ci(−≤xi,ci≤), 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 InputSample OutputSource
2017中国大学生程序设计竞赛 - 女生专场
题目大意:有N个教室,需要在教室建糖果店,所需要的花费有两种情况
1.建糖果店所需要花费的费用。
2.这个点离左边最近的糖果店的距离。
分析:这个点有两种可能 建糖果店 不建糖果店 所以是道dp题
设dp【i】【0】表示在这个点建立糖果店所需要花费的最小费用,dp【i】【0】在这个点不建糖果店所需要花费的最小费用
dp【i】【1】=min(dp[i-1][1],dp[i-1][0])+ci;
dp[i][0]的情况需要从起点到开始枚举,所以时间复杂度为O(n*n);
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include <vector>
#include<iostream>
using namespace std;
#define N 50005
#define INF 0x3f3f3f3f
#define LL long long
LL dp[N][];
struct node
{
LL x,c;
}s[N];
int cmp(node a,node b)
{
return a.x<b.x;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=;i<=n;i++)
scanf("%lld %lld",&s[i].x,&s[i].c);
sort(s+,s+n+,cmp);
memset(dp,,sizeof(dp));
dp[][] = s[].c;///东边肯定有糖果店 所以第一个教室是必须建糖果店的
dp[][] = INF;
for(int i=;i<=n;i++)
{
dp[i][] = min(dp[i-][],dp[i-][])+s[i].c;
///在这个教室建立糖果店所需要的最小花费
LL sum=;
dp[i][] = INF;
///找到前面花费最小的糖果店
for(int j=i-;j>=;j--)
{
sum+=(i-j)*(s[j+].x-s[j].x);
dp[i][] = min(dp[i][],dp[j][]+sum);
}
}
printf("%lld\n",min(dp[n][],dp[n][]));
///取建糖果店与不建糖果店的其中的小值
}
return ;
}