POJ3253 Frence repair

Fence Repair

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 47901 Accepted: 15734

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

解题思路：最小堆:

AC情况:

AC的代码：

# include <stdio.h># define N 20000void insert(int Tree[],int X)//向Tree堆里插入X{    int par,i=++Tree[0];  //插入X 后 Tree[0]+1    while(i>1)  //直到i不是根节点    {       par=i/2;  //父节点为par       if(Tree[par]<=X) break;//如果父节点满足最大堆的特性 则插入当前的位置即可       Tree[i]=Tree[par]; //否则调整堆 即位置上移       i=par;    }    Tree[i]=X;//插入新节点}int DelMin(int Tree[])//删除最大值{   int i=1,root=Tree[1],R,L,X=Tree[Tree[0]--];//X记录当前末尾节点 root为根 即为最大值   if(Tree[0]<0)  return -1;   while(i*2<=Tree[0])   {      L=i*2;  R=L+1;//Left Right 记录左右节点      if(R<=Tree[0]&&Tree[R]<Tree[L])L=R;      if(Tree[L]>=X) break;//如果所有的位置已经调整好 跳出循环      Tree[i]=Tree[L];//否则继续调整堆      i=L;   }   Tree[i]=X;   return  root;}int L[N+1],i,n,A;//L数组表示堆long long sum;int main(){    scanf("%d",&n);    for(i=0;i<n;i++)    {        scanf("%d",&A);        insert(L,A);    }    while(L[0]>1)    {        A=DelMin(L)+DelMin(L);        sum+=A;        insert(L,A);    }    printf("%lld\n",sum);    return 0;}

代码2：

# include <stdio.h># define N 20000void insert(int Tree[],int X)//向Tree堆里插入X{    int par,i=++Tree[0];  //插入X 后 Tree[0]+1    while(i>1)  //直到i不是根节点    {       par=i>>1;  //父节点为par       if(Tree[par]<=X) break;//如果父节点满足最大堆的特性 则插入当前的位置即可       Tree[i]=Tree[par]; //否则调整堆 即位置上移       i=par;    }    Tree[i]=X;//插入新节点}int DelMin(int Tree[])//删除最大值{   int i=1,root=Tree[1],R,L,X=Tree[Tree[0]];//X记录当前末尾节点 root为根 即为最大值   if(Tree[0]<0)  return -1;   while(i<<1<Tree[0]){      L=i<<1;  R=L+1;//Left Right 记录左右节点      if(R<Tree[0]&&Tree[R]<Tree[L])L=R;      if(Tree[L]>=X) break;//如果所有的位置已经调整好 跳出循环      Tree[i]=Tree[L];//否则继续调整堆      i=L;   }   Tree[i]=X;   Tree[0]--;   return  root;}int L[N+1],n,A;//L数组表示堆long long sum;int main(){    scanf("%d",&n);    while(n--){        scanf("%d",&A);        insert(L,A);    }    while(L[0]>1)    {        A=DelMin(L)+DelMin(L);        sum+=A;        insert(L,A);    }    printf("%lld\n",sum);    return 0;}