Building Shops
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2659 Accepted Submission(s): 911
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.
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.
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
3 1 2 2 3 3 4 4 1 7 3 1 5 10 6 1
Sample Output
5 11
Source
Recommend
jiangzijing2015
这题一看就是一个动态规划题,题目意思为你有n个教室,要满足以下的情况
1如果当前教室要造糖果点,费用要加上该店的费用
2如果当前教室不造糖果点,则费用为它到它左端最近糖果点的路程差(第一个教室一定有糖果点)
二维dp,dp[i][j],i代表classroom数量,j代表下标最大糖果点的位置
还有一点,数据不是按照x坐标升序给出的,也就是说要先对x排序,我没排序wa了2-3次,还以为是其他问题
#include<cstdio>
#include<algorithm>
using namespace std;
long long dp[3005][3005];
struct me{
long long s,v;
}my[3005];
bool cmp(struct me a,struct me b){
return a.s<b.s;
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++)
scanf("%lld%lld",&my[i].s,&my[i].v);
sort(1+my,my+1+n,cmp);
dp[1][1]=my[1].v;
for(int i=2;i<=n;i++)
{
for(int j=1;j<i;j++)
dp[i][j] = dp[i-1][j] + my[i].s - my[j].s;
dp[i][i] = 1e18;
for(int j=1;j<i;j++)
dp[i][i] = min(dp[i][i], dp[i-1][j]+my[i].v);
}
long long mmax=1e18;
for(int i=1;i<=n;i++)
mmax=min(mmax,dp[n][i]);
printf("%lld\n",mmax);
}
return 0;
}