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UVA 10954- Add All(優先隊列)

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UVA 10954- Add All(優先隊列)

Add All Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status

Description

Problem F
Add All
Input: standard input
Output: standard output

Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply question your erudition. So, let’s add some flavor of ingenuity to it.

Addition operation requires cost now, and the cost is the summation of those two to be added. So, to add 1 and 10, you need a cost of 11. If you want to add 1, 2 and 3. There are several ways –

1 + 2 = 3, cost = 3

3 + 3 = 6, cost = 6

Total = 9

1 + 3 = 4, cost = 4

2 + 4 = 6, cost = 6

Total = 10

2 + 3 = 5, cost = 5

1 + 5 = 6, cost = 6

Total = 11

I hope you have understood already your mission, to add a set of integers so that the cost is minimal.

Input

Each test case will start with a positive number, N (2 ≤ N ≤ 5000) followed by N positive integers (all are less than 100000). Input is terminated by a case where the value of N is zero. This case should not be processed.

Output

For each case print the minimum total cost of addition in a single line.

Sample Input Output for Sample Input

1 2 3

1 2 3 4

Problem setter: Md. Kamruzzaman, EPS

題意：有n個數的集合s，每次可以從s中刪除兩個數，然後把他們的和放回集合，直至剩下一個數。每次的開銷等於刪除的兩個數的和，求最小總開銷。

思路：使用從小到大排序的優先隊列。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int inf=0x3f3f3f3f;
int main()
{
    int n,i,a;
    int sum;
    while(~scanf("%d",&n)){
        if(n==0) break;
        priority_queue,greater > q;
        while(n--){
            scanf("%d",&a);
            q.push(a);
        }
        sum=0;
        while(!q.empty()){
            int x=q.top();
            q.pop();
            int y=q.top();
            q.pop();
            sum+=x+y;
            if(!q.empty())
            q.push(x+y);
        }
        printf("%d\n",sum);
    }
}