Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23913		Accepted: 7595

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length L_i (1 ≤ L_i ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths L_i). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.

FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer 

N, the number of planks 

Lines 2..

N+1: Each line contains a single integer describing the length of a needed plank

Output

Line 1: One integer: the minimum amount of money he must spend to make 

N-1 cuts

Sample Input

3858

Sample Output

34

Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8. 

The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

Source

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　　堆排序应用，简单题。

　　思路是先用堆排序找到最短的一块木板，记下来，删除它，再对长度减1的数组堆排序，将堆顶最小的元素和刚才最小的相加，这就以最小的代价合成了第一块木板（贪心），之后不断循环这个过程直到合成到只有一块木板。最后输出代价之和。

　　直接套模板过。没搞透，有时间再好好看看。

　　堆排序模板：

#define maxn 20000+2int l[maxn];void heap_sort(int h,int x){    int lg,lr,t;    while((x<<1) <= h){        lg = x<<1;        lr = (x<<1) + 1;        if( lr<=h && l[lg] > l[lr])            lg = lr;        if( l[x] > l[lg] ){            t = l[x];            l[x] = l[lg];            l[lg] = t;            x = lg;        }        else             break;    }}

　　代码：

1 #include 
      
        2 #define maxn 20000+2 3 int l[maxn]; 4 void heap_sort(int h,int x) 5 { 6     int lg,lr,t; 7     while((x<<1) <= h){ 8         lg = x<<1; 9         lr = (x<<1) + 1;10         if( lr<=h && l[lg] > l[lr])11             lg = lr;12         if( l[x] > l[lg] ){13             t = l[x];14             l[x] = l[lg];15             l[lg] = t;16             x = lg;17         }18         else 19             break;20     }21 }22 int main()23 {24     int i,n;25     while(scanf("%d",&n)!=EOF){26         int Min;27         long long ans = 0;28         for(i=1;i<=n;i++)    //输入29             scanf("%d",&l[i]);30         //建堆31         for(i=n/2;i>0;i--)32             heap_sort(n,i);33         //work34         while(n>1){35             heap_sort(n,1);36             Min = l[1];37             l[1] = l[n--];38             heap_sort(n,1);39             Min += l[1];40             l[1] = Min;41             ans+=Min;42         }43         printf("%I64d\n",ans);44     }45     return 0;46 }
      

 

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