其实本题可以水过
var a:array[0..10000] of longint;
n,total:longint;
i:integer;

procedure swap(var a,b:longint);
var t:integer;
begin
t:=a; a:=b; b:=t
end;

procedure quick_sort(m,n:integer);
var i,j,x:integer;
begin
i:=m; j:=n; x:=a[(i+j) div 2];
repeat
while a[i]>x do inc(i);
while x>a[j] do dec(j);
if i<=j then begin
swap(a[i],a[j]);
inc(i); dec(j)
end;
until i>j;
if m<j then quick_sort(m,j);
if i<n then quick_sort(i,n);
end;

procedure insert_sort(n:integer);
var i,j:integer;
begin
i:=n-1; a[0]:=a[n];
while a[i]<a[0] do begin
a[i+1]:=a[i];
dec(i)
end;
a[i+1]:=a[0]
end;

begin
readln(n);
for i:=1 to n do
read(a[i]);
quick_sort(1,n);
for i:=1 to n-1 do begin
a[n-i]:=a[n-i]+a[n-i+1];
a[n-i+1]:=0;
inc(total,a[n-i]);
insert_sort(n-i)
end;
writeln(total);
end.

/* ***********************************************
Author        :guanjun
Created Time  :2016/2/19 8:40:51
File Name     :vijosp1097.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
struct item{
    int v;
    bool operator <(const item &a)const{
        return v>a.v;
    }
};

bool cmp(int a,int b){
    return a>b;
}
priority_queue<item>pq;
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n;
    while(cin>>n){
        int a;
        for(int i=1;i<=n;i++){
            cin>>a;
            item w;
            w.v=a;
            pq.push(w);
        }
        ll ans=0;           
        item x,y,z;
        int sum=0;
        while(--n){
            x=pq.top();pq.pop();
            y=pq.top();pq.pop();
            sum=x.v+y.v;
            z.v=sum;
            pq.push(z);
            ans+=sum;
        }
        cout<<ans<<endl;
    }
    return 0;
}

#include<stdio.h>
#include<queue>
using namespace std;
priority_queue<int,vector<int>,greater<int> > q;
int n,ans=0;
int main(){
    scanf("%d",&n);
    for(int x,i=0;i<n;++i){
        scanf("%d",&x);
        q.push(x);
    }
    while(--n){
        int x=q.top(); q.pop();
        int y=q.top(); q.pop();
        ans+=x+y;
        q.push(x+y);
    }
    printf("%d",ans);
}

优先队列解题无压力，讨论版上的哈夫曼树不过也是醉了
#include<iostream>
#include<cstdio>
#include<functional>
#include<queue>
using namespace std;
priority_queue<int, vector<int>, greater<int> >guozi;
int m;
int a,b;
int main(){
scanf("%d",&m);
for(int i=1;i<=m;i++){
scanf("%d",&a);
guozi.push(a);
}
int sum=0;
while(guozi.size()>1){
sum=sum+guozi.top();
b=guozi.top();

guozi.pop();

sum=sum+guozi.top();
int k=guozi.top();
guozi.pop();
guozi.push(b+k);
}
printf("%d",sum);
return 0;
}

评测结果
编译成功

foo.cpp: In function 'int main()':
foo.cpp:16:19: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
if(k2==tot||k1<n&&a[k1+1]<b[k2+1])
^
foo.cpp:19:19: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
if(k2==tot||k1<n&&a[k1+1]<b[k2+1])
^
测试数据 #0: Accepted, time = 15 ms, mem = 644 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 644 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 644 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 644 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 644 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 644 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 640 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 644 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 644 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 640 KiB, score = 10
Accepted, time = 105 ms, mem = 644 KiB, score = 100

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[10003],b[10003];
int main()
{
int n,ans=0,x=0;
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i];
sort(a+1,1+a+n);
int k1=0,k2=0,tot=0;
for(int i=1;i<=n-1;i++)
{
if(k2==tot||k1<n&&a[k1+1]<b[k2+1])
x=a[++k1];
else x=b[++k2];
if(k2==tot||k1<n&&a[k1+1]<b[k2+1])
x+=a[++k1];
else x+=b[++k2];
ans+=x;
b[++tot]=x;
}
cout<<ans;
}

手写堆，还是挺快的
记录信息
评测状态 Accepted
题目 P1097 合并果子
递交时间 2015-11-22 10:05:00
代码语言 C++
评测机 VijosEx
消耗时间 45 ms
消耗内存 316 KiB
评测时间 2015-11-22 10:05:02
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 316 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 316 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 316 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 316 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 312 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 312 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 312 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 316 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 312 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 316 KiB, score = 10
Accepted, time = 45 ms, mem = 316 KiB, score = 100
代码
#include <iostream>
#include <stdio.h>
using namespace std;
int heap[10010],cnt;
void add(int x)
{ int p=++cnt;
while(p!=1)
{
int t=p/2;
if(heap[t]<=x)break;
heap[p]=heap[t];
p=t;
}
heap[p]=x;
}
void del()
{ if(cnt==0)return;
int p=1,temp=heap[cnt--];
while(1)
{ int a=p*2,b=p*2+1;
if(a>cnt)break;
if(heap[a]>heap[b]&&b<=cnt)a=b;
if(temp<=heap[a])break;
heap[p]=heap[a];
p=a;
}
heap[p]=temp;
}
int main()
{ int n,ans=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{ int a;
scanf("%d",&a);
add(a);
}
for(int i=1;i<n;i++)
{ int a=heap[1];del();
int b=heap[1];del();
ans+=a+b;
add(a+b);
}
printf("%d",ans);
return 0;
}

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int n,x,y,z,ans;
priority_queue<int,vector<int>,greater<int> >q;
int main()
{
cin>>n;
for(int i=1;i<=n;i++){
cin>>x;q.push(x);
}
ans=0;
while(q.size()>=2){
y=q.top();q.pop();
z=q.top();q.pop();
x=y+z;ans=ans+x;
q.push(x);
}
x=q.top();
ans=ans;
cout<<ans;
}
其实用优先队列最简单。。。。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

int n,i,j,a[10002],ans,tot;

int comp(int a,int b)

{

if(a<b) return 1;

else return 0;

}

int main()

{

cin>>n;

for(i=1;i<=n;i++) cin>>a[i];

sort(a+1,a+n+1,comp);

for(i=1;i<n;i++)
{
ans=a[i]+a[i+1];
a[n+1]=ans;
tot=tot+ans;
for(j=i+1;j<=n;j++)
{
a[j]=a[j+1];
if(a[j+1]>ans) { a[j]=ans; break;}

}

}

cout<<tot;

return 0;

}

用的最简单的方法，不过AC了……好高兴呢~
program combine;
var
n,m,i,min1i,min2i,num,mun:integer;
min1,min2,sum:longint;
a:array[1..20000] of longint;
begin
readln(n);
for i:=1 to n do
read(a[i]);
min1:=maxlongint;
min2:=maxlongint;
m:=n;
sum:=0;
repeat
for i:=1 to m do
if a[i]<min1 then
begin
min1:=a[i];
min1i:=i;
end;
for i:=1 to m do
if (a[i]<min2) and (a[i]>=min1) and (i<>min1i) then
begin
min2:=a[i];
min2i:=i;
end;
if min1i<min2i then
begin
num:=min1i;
mun:=min2i;
end
else
begin
num:=min2i;
mun:=min1i;
end;
a[num]:=min1+min2;
for i:=mun to m-1 do
a[i]:=a[i+1];
a[m]:=0;
sum:=sum+min1+min2;
min1:=maxlongint;
min2:=maxlongint;
m:=m-1;
until m=1;
write(sum);
end.

1、调用库，37行秒杀
2、队列，3种情况判断
3、堆（练手专用

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int n;
int i;
int j;
vector<int>apple;

void llo(int f)
{
if(f==n-1)return;
if(apple[f]>apple[f+1])
{

j=apple[f];
apple[f]=apple[f+1];
apple[f+1]=j;
llo(f+1);
}
else return;
}

int main()
{
cin>>n;
int he[n-1];
for(i=0;i<n;++i)
{
cin>>j;
apple.push_back(j);
}
sort(apple.begin(),apple.end());
long int w=0;
int f=0;
for(i=0;i<n-1;i++)
{
j=apple[f]+apple[f+1];
++f;
apple[f]=j;
w+=apple[f];
llo(f);
}
cout<<w;
return 0;
}

AC 12 ti liu nian !!!!!!!!!!!!!!!!!!!!!!!!!!!!!! hahahaahahahahaha !!!!!!!!!!!!!!!!!!!!!!!!!!!!

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;

int a[10010];

void change(int &a, int&b){
int t = a;
a = b;
b = t;

}

int main()
{
int n;
long long ans = 0;
scanf("%d", &n);
for(int i=1; i<=n; i++)
scanf("%d", &a[i]);
sort(&a[1], &a[1]+n);
for(int i=1; i<n; i++){
int j = i;
while(a[j] > a[j+1]){
if(j >= n)
break;
change(a[j], a[j+1]);
j++;
}
ans += a[i] + a[i+1];
a[i+1] += a[i];
}
printf("%lld", ans);
system("pause");
return 0;
}
注意每次可以在原排序的基础上修改

#include <iostream>
#include <algorithm>
using namespace std;
int ans,n;
int fruit[10005];
void input()
{
cin>>n;
for(int i=1;i<=n;i++)cin>>fruit[i];
return;
}
int main()
{
for(int i=1;i<=10004;i++)fruit[i]=2147483647;//初始化 防止诡异的越界
input();//输入数据
sort(fruit+1,fruit+1+n);//先排好升序
for(int pos=1;pos<=n-1;pos++)
{
int temp=pos;
while(fruit[temp]>fruit[temp+1])//重新排序的过程
{
swap(fruit[temp],fruit[temp+1]);
temp++;
}
fruit[pos+1]+=fruit[pos];//合并到下一堆里去
ans+=fruit[pos+1];//耗费体力
}
cout<<ans;
return 0;
}

#include<iostream>
using namespace std;
main()
{
int z[10000],t,i,w,j[10000];
cin>>i;
for(int a=1;a<=i;a++)
{
cin>>z[a];
}
for(int a=1;a<=i;a++)
{
for(int b=1;b<=i;b++)
{
if(z[a]>=z[b])
{
t=z[b];
z[b]=z[a];
z[a]=t;
}
}
}
for(int a=i;a>=1;a--)
{

w=a;
for(int b=w-2;b>=1;b--)
{
if(z[b]<=z[a-1])
{
t=z[b];
z[b]=z[a-1];
z[a-1]=t;
}
}
z[a-1]=z[a]+z[a-1];
z[a]=0;
j[a-1]=z[a-1];
}
t=0;
for(int a=1;a<=i-1;a++){
t=j[a]+t;
}
cout<<t;
}

#include <iostream>
#include <algorithm>
#define MAXN 10001
using namespace std;
int heap_size=0;
int heap[MAXN],input[MAXN];
/*
void put(int num){
int now,next;
heap[++heap_size]=num;
now=heap_size;
while(now>1){
next=now>>1;
if(heap[now]>=heap[next]) break;
swap(heap[now],heap[next]);
now=next;
}
}
void get(){
int now,next;
heap[1]=heap[heap_size];
heap_size--;
now=1;
while(now*2<=heap_size){
next=now*2;
if(next<heap_size && heap[next+1]<heap[next]) next++;
if(heap[now]<=heap[next]) break;
swap(heap[now],heap[next]);
now=next;
}
}
*/
void put(int num){
heap[++heap_size]=num;
push_heap(heap+1,heap+heap_size+1,greater<int>());
}
int get(){
int res;
res=heap[1];
pop_heap(heap+1,heap+heap_size+1,greater<int>());
heap_size--;
return res;
}
int main(){
int n,ans=0;
cin>>n;
for(int i=1;i<=n;i++){
cin>>input[i];
put(input[i]);
}
while(heap_size>1){
int tmp1=get();
int tmp2=get();
int tmp=tmp1+tmp2;
ans+=tmp;
put(tmp);
}
cout<<ans<<endl;
return 0;
}
/*
10
3 5 1 7 6 4 2 5 4 1
*/

请记住，我叫傻蛋，抄了我的程序并AC后，请记得回踩哦！！！！！
-----我是傻蛋 2015/9/1

这道题呢其实是用贪心算法，先寻找最小的两堆合并，但注意每合并一次都要重新排序。数据还是不大的，用long long能过。
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n;
void insort(long w[],int x)
{
bool bo;
int i,temp;
for(i=x;i<=n-1;i++){
bo=true;
if(w[i]>w[i+1]){
temp=w[i];
w[i]=w[i+1];
w[i+1]=temp;
bo=false;
}
if(bo==true){
break;
}
}
}
int main()
{
bool bo;
long a[10001],sum[10001]={0};
int i,tail=1,j,temp;;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
}
for(i=1;i<=n-1;i++){
do{
bo=true;
for(j=1;j<=n-i;j++){
if(a[j]>a[j+1]){
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
bo=false;
}

}
}while(bo==false);
if(bo==true){
break;
}
}
for(i=2;i<=n;i++){
sum[tail]=a[i]+a[i-1];
a[i]=sum[tail];
insort(a,i);
tail++;
}
for(i=2;i<tail;i++){
sum[i]=sum[i-1]+sum[i];
}
printf("%d",sum[tail-1]);
return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;
int n,a[20001];
long long ans=0;
int main()
{
int temp,jj;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
int minn;
for(int g=1;g<n;g++)
for(int i=0;i<=1;i++)
{
minn=999999999;
for(int j=g+i;j<=n;j++)
if(a[j]<minn) minn=a[j],jj=j;
temp=a[g+i],a[g+i]=a[jj],a[jj]=temp;
ans+=minn;
if(i==1) a[g+1]+=a[g];
}
printf("%I64d\n",ans);
return 0;
}

var heap:array[1..10000] of longint;
i,num,n,size,a,b:longint;
ans:int64;
procedure put(x:longint);
var son,fa,t:longint;
begin
inc(size); heap[size]:=x; son:=size;
fa:=son div 2;
while (son<>1) and (heap[fa]>heap[son]) do
begin
t:=heap[fa]; heap[fa]:=heap[son]; heap[son]:=t;
son:=fa; fa:=son div 2;
end;

end;
function get:longint;
var fa,son,t:longint;
begin
get:=heap[1]; heap[1]:=heap[size]; dec(size); fa:=1;
while (fa*2<=size) do
begin
if (fa*2+1>size) or (heap[fa*2]<heap[fa*2+1]) then
son:=fa*2
else
son:=fa*2+1;
if heap[fa]>heap[son] then
begin
t:=heap[fa]; heap[fa]:=heap[son]; heap[son]:=t; fa:=son;
end
else
break;
end;
end;

begin //main
read(n); size:=0; fillchar(heap,sizeof(heap),0); ans:=0;
for i:=1 to n do
begin
read(num); put(num);
end;

while size<>1 do
begin
a:=get; b:=get;
ans:=a+b+ans; a:=a+b; put(a);
end;

write(ans);
end.