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  • HDU 6024 Building Shops(DP)

    Building Shops

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 889    Accepted Submission(s): 343


    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(1n3000), denoting the number of the classrooms.
    In the following n lines, each line contains two integers xi,ci(109xi,ci109), 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

    2017中国大学生程序设计竞赛 - 女生专场



    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstdlib>
    #include<queue>
    #include<map>
    #include<set>
    #include<stack>
    #include<bitset>
    #include<numeric>
    #include<vector>
    #include<string>
    #include<iterator>
    #include<cstring>
    #include<ctime>
    #include<functional>
    #define INF 0x3f3f3f3f
    #define ms(a,b) memset(a,b,sizeof(a))
    #define pi 3.14159265358979
    #define MOD 1000000007
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    using namespace std;
    
    typedef pair<int, int> P;
    typedef long long ll;
    typedef unsigned long long ull;
    const int maxn = 3010;
    
    struct Node {
    	ll x, c;
    }a[maxn];
    
    ll dp[maxn][maxn];
    
    bool cmp(Node a, Node b)
    {
    	return a.x < b.x;
    }
    
    int main()
    {
    		int n;
    		while (~scanf("%d", &n))
    		{
    			for (int i = 1; i <= n; i++)
    			{
    				scanf("%lld%lld", &a[i].x, &a[i].c);
    			}
    			sort(a + 1, a + 1 + n, cmp);
    			ms(dp, INF);
    			dp[1][1] = a[1].c;
    			for (int i = 2; i <= n; i++)
    			{
    				for (int j = 1; j < i; j++)
    				{
    					dp[i][i] = min(dp[i][i], dp[i - 1][j] + a[i].c);
    					dp[i][j] = min(dp[i][j], dp[i - 1][j] + (a[i].x - a[j].x));
    				}
    			}
    			ll ans = INF;
    			for (int i = 1; i <= n; i++)
    			{
    				ans = min(ans, dp[n][i]);
    			}
    			printf("%lld
    ", ans);
    
    		}
    		
    }



    Fighting~
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  • 原文地址:https://www.cnblogs.com/Archger/p/12774773.html
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