题解

535 条题解

  • 0
    @ 2016-08-24 20:42:02

    用c++stl库就行啦

    #include<cstdio>
    #include<iostream>
    #include<queue>
    using namespace std;
    long long n,x,y,ans=0;
    priority_queue<long long,vector<long long>,greater<long long> > h;
    int main()
    {
    scanf("%lld",&n);
    for(int i=1;i<=n;i++)
    {
    scanf("%lld",&x);
    h.push(x);
    }
    for(int i=1;i<=n-1;i++)
    {
    x=h.top();h.pop();
    y=h.top();h.pop();
    ans+=x+y;h.push(x+y);
    }
    printf("%lld",ans);
    return 0;
    }

  • 0
    @ 2016-08-16 10:16:10

    program fruit;
    var
    a:array [1..10000] of longint;
    i,m,n,x,j,total:longint;

    procedure QuickSort(low,high:longint);
    var
    i,j,mid,med:longint;
    begin
    i:=low;
    j:=high;
    mid:=a[(i+j) div 2];
    repeat
    while a[i]<mid do
    i:=i+1;
    while a[j]>mid do
    j:=j-1;
    if i<=j then
    begin
    med:=a[i];
    a[i]:=a[j];
    a[j]:=med;
    i:=i+1;
    j:=j-1;
    end;
    until i>j;
    if i<high then QuickSort(i,high);
    if j>low then QuickSort(low,j);
    end;

    begin
    readln(n);
    for i:=1 to n do
    read(a[i]);
    QuickSort(1,n);
    for i:=1 to n-1 do
    begin
    x:=a[i]+a[i+1];
    a[i]:=0;
    a[i+1]:=0;
    total:=total+x;
    j:=i+2;
    while (j<=n) and (x>a[j]) do
    j:=j+1;
    j:=j-1;
    for m:=i+1 to j do
    a[m]:=a[m+1];
    a[j]:=x;
    end;
    write(total);
    end.

  • 0
    @ 2016-08-15 00:30:02
    #include <bits/stdc++.h>
    #define maxN 100010
    
    using namespace std;
    typedef long long QAQ ;
    
    int arr[maxN];
    
    priority_queue<int ,vector<int>,greater<int> >Q; 
    QAQ ans ;
    int main(){
        int n,tmp2,tmp1;
        ios::sync_with_stdio(false);
        cin >> n ;
        for(int i=1 ; i<=n ;++i)cin >> arr[i];
        sort(arr+1,arr+n+1);
        for(int i=1;i<=n;++i){
            Q.push(arr[i]);
        }
        for(int i=1;i<=n-1;++i){
            tmp1=Q.top();
            Q.pop();
            tmp2=Q.top();
            Q.pop();
            Q.push(tmp1+tmp2);
            ans+=tmp1+tmp2;
        }
        printf("%lld",ans);
        return 0 ;
    } 
    
  • 0
    @ 2016-08-13 15:05:20

    !最小生成树+排序算法解决:
    #include <iostream>
    using namespace std;
    void run(int*pData,int left,int right)
    {
    int i,j;
    int middle,iTemp;
    i=left;
    j=right;
    middle=pData[(left+right)/2];//求中间值
    do
    {
    while((pData[i]<middle)&&(i<right))//从左扫描小于中值的数
    i++;
    while((pData[j]>middle)&&(j>left))//从右扫描大于中值的数
    j--;
    if(i<=j)//找到了一对值
    {
    //交换
    iTemp=pData[i];
    pData[i]=pData[j];
    pData[j]=iTemp;
    i++;
    j--;
    }
    }while(i<=j);//如果两边扫描的下标交错,就停止(完成一次)

    //当左边部分有值(left<j),递归左半边
    if(left<j)
    run(pData,left,j);
    //当右边部分有值(right>i),递归右半边
    if(right>i)
    run(pData,i,right);
    }
    void QuickSort(int*pData,int Count)
    {
    run(pData,0,Count-1);
    }

    int main()
    {
    int n,a[10000],i,j,k,t,is,sum=0;
    cin>>n;
    for(i=0;i<n;i++)
    {
    cin>>a[i];
    }
    QuickSort(a,n);
    for(i=0;i<n-1;i++)
    {
    sum+=a[i]+a[i+1];
    a[i+1]=a[i]+a[i+1];
    is=0;
    for(j=i+2;j<n;j++)
    {
    if(a[j]>=a[i+1])
    {
    is=1;
    t=a[i+1];
    for(k=i+1;k<j-1;k++)a[k]=a[k+1];
    a[j-1]=t;
    break;
    }
    }
    if(is==0)
    {
    t=a[i+1];
    for(k=i+1;k<n-1;k++)a[k]=a[k+1];
    a[n-1]=t;
    }
    }

    cout<<sum<<endl;

    return 0;
    }

  • 0
    @ 2016-08-08 15:10:48

    哈夫曼树
    #include <stdio.h>
    #include <string.h>
    struct node
    {
    int father;
    int weight;
    int old;
    };
    node tree[100000];
    int height(int h1)
    {
    if (tree[h1].father==-1) return 0;
    return height(tree[h1].father)+1;
    }
    int main()
    {
    int n,tot;
    scanf("%d",&n);
    tot=n;
    for (int i=1;i<=n;i++)
    {
    scanf("%d",&tree[i].weight);
    tree[i].old=1;
    }
    for (int i=1;i<=2*n-1;i++)
    tree[i].father=-1;
    for (int i=1;i<=2*n-2;i++)
    {
    int p1=100000000;
    int k1=0;
    for (int j=1;j<=2*n-1;j++)
    if (tree[j].father==-1 && tree[j].weight<p1 && tree[j].weight)
    {
    k1=j;
    p1=tree[j].weight;
    }
    int p2=1000000000;
    int k2=0;
    for (int j=1;j<=2*n-1;j++)
    if (tree[j].father==-1 && tree[j].weight<p2 && tree[j].weight && j!=k1)
    {
    k2=j;
    p2=tree[j].weight;
    }
    if (!k1 || !k2) break;
    int k3=0;

    if (k3==0)
    {
    tot=tot+1;
    k3=tot;
    }
    tree[k3].weight=tree[k2].weight+tree[k1].weight;
    tree[k2].father=tree[k1].father=k3;
    }
    int add=0;
    for (int i=1;i<=2*n-1;i++)
    if (tree[i].old)
    add=add+tree[i].weight*(height(i));
    printf("%d",add);
    return 0;
    }

  • 0
    @ 2016-07-26 06:55:08

    紫书例题。。

    #include<iostream>
    #include<queue>

    int main()
    {
    using namespace std;
    int ans=0,n,o;
    priority_queue<int,vector<int>,greater<int> >q;
    cin>>n;
    for(int i = 0; i < n; i++ ){
    cin>>o;q.push(o);}
    for(int i = 0;i < n-1 ; i++)
    {
    int a = q.top();q.pop();
    int b = q.top();q.pop();
    ans+=a+b;
    q.push(a+b);
    }
    cout<<ans;
    return 0;
    }

  • 0
    @ 2016-07-22 21:12:01
    #include<iostream>
    #include<cstdio>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    int n;
    
    int main ()
    {
        priority_queue<int, vector<int>, greater<int> > Q;
        cin >> n;
        int T = n - 1;
        int x;
        while (n--) { cin >> x; Q.push(x);}
        int ans = 0;
        while (T--) {
            int w1 = Q.top(); Q.pop();
            int w2 = Q.top(); Q.pop();
            ans += w1 + w2;
            Q.push(w1+w2);
        }
        cout << ans << "\n";
        return 0;
    }
    
  • 0
    @ 2016-07-22 21:00:09

    var i,j,x,n,max:longint;
    a:array[1..100000000] of longint;
    flag:boolean;
    ans:int64;
    begin
    ans:=0;max:=0;
    readln(n);fillchar(a,sizeof(a),0);
    for i:=1 to n do
    begin
    read(x);
    inc(a[x]);
    max:=max+x;
    end;
    if max>100000000 then max:=100000000;
    i:=1;
    while i<=max do
    begin
    flag:=false;
    if a[i]>=2 then
    repeat
    a[i]:=a[i]-2;
    inc(a[2*i]);
    ans:=ans+2*i;
    until a[i]<2;
    if a[i]=1 then
    for j:=i+1 to max do
    begin
    if (a[i]>0) and (a[j]>0) then
    begin
    dec(a[i]);
    dec(a[j]);
    inc(a[i+j]);
    ans:=ans+i+j;
    flag:=true;
    end;
    if flag then break;
    end;
    inc(i);
    end;
    write(ans);

    end.

  • 0
    @ 2016-07-14 16:55:24

    优先队列

    #include <iostream>
    #include <cstdio>
    #include <queue>
    using namespace std;
    struct data{
        int val;
        inline bool operator < (const data&data1) const{
            return data1.val<val;
        }
    };
    template<class T>
    class PriorityQueue : public priority_queue<T>{
    public:
        T pop (){
            T tmp=priority_queue<T>::top();
            priority_queue<T>::pop();
            return tmp;
        }
    };
    PriorityQueue <data> que;
    int main(int argc, char const *argv[]){
        /* code */
        //freopen("165.in","r",stdin);
        //freopen("165.out","w",stdout);
        ios::sync_with_stdio(false);
        cin.tie(NULL);
        int n,ans=0;
        cin>>n;
        data tmp;
        for(int i=0;i<n;i++){
            cin>>tmp.val;
            que.push(tmp);
        }
        for(int i=1;i<n;i++){
            tmp=que.pop();
            tmp.val+=que.pop().val;
            ans+=tmp.val;
            que.push(tmp);
        }
        cout<<ans;
        return 0;
    }
    
  • 0
    @ 2016-07-14 10:22:36

    #include <cstdio>
    #include <queue>

    using std::priority_queue;
    using std::vector;
    using std::greater;

    int main(){
    int n,sum=0,t;
    priority_queue<int,vector<int>,greater<int> > q;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
    scanf("%d",&t);
    q.push(t);
    }
    for(int i=1;i<=n-1;i++){
    int a1=q.top();
    q.pop();
    int a2=q.top();
    q.pop();
    sum+=a1+a2;
    q.push(a1+a2);
    }
    printf("%d",sum);
    return 0;
    }

  • 0
    @ 2016-07-10 13:32:13

    测试数据 #0: Accepted, time = 140 ms, mem = 860 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 864 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 860 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 856 KiB, score = 10
    测试数据 #4: Accepted, time = 31 ms, mem = 864 KiB, score = 10
    测试数据 #5: Accepted, time = 31 ms, mem = 864 KiB, score = 10
    测试数据 #6: Accepted, time = 125 ms, mem = 864 KiB, score = 10
    测试数据 #7: Accepted, time = 125 ms, mem = 860 KiB, score = 10
    测试数据 #8: Accepted, time = 125 ms, mem = 860 KiB, score = 10
    测试数据 #9: Accepted, time = 140 ms, mem = 864 KiB, score = 10
    Accepted, time = 717 ms, mem = 864 KiB, score = 100

    var a:array[0..15000] of longint;
    i,j,n,k,tem,sum:longint;

    procedure init;
    begin
    readln(n);
    for i:=1 to n do
    read(a[i]);
    end;

    procedure qsort(left,right:integer);

    var i,j,x,t:longint;
    begin
    i:=left; j:=right;
    x:=a[i];
    repeat
    while (a[j]<x)and(i<j) do dec(j);
    if i<j then
    begin
    t:=a[i]; a[i]:=a[j];a[j]:=t;inc(i);
    end;
    while (a[i]>x)and(j>i) do inc(i);
    if i<j then
    begin
    t:=a[i]; a[i]:=a[j];a[j]:=t;dec(j);
    end;
    until i=j;
    inc(i);dec(j);
    if i<right then qsort(i,right);
    if j>left then qsort(left,j);
    end;

    procedure work;
    begin
    sum:=0;
    while n>=2 do
    begin
    tem:=a[n]+a[n-1];
    sum:=sum+tem;
    j:=1;
    while a[j]>tem do j:=j+1;
    for k:=n downto j do
    a[k+1]:=a[k];
    a[j]:=tem;
    n:=n-1;
    end;
    end;

    begin
    init;
    qsort(1,n);
    work;
    writeln(sum);
    end.

  • 0
    @ 2016-06-24 10:38:28

    #include<cstdio>
    int heap[30001];
    int n;
    void swap(int &a,int &b){
    int t=b;
    b=a;
    a=t;
    }
    int len;
    void put(int val){
    int now,next;
    heap[++len]=val;
    now=len;
    while(now>1){
    int next=now/2;
    if(heap[next]<=heap[now]){
    return;
    }
    else{
    swap(heap[next],heap[now]);
    now=next;
    }
    }
    }
    int get(){
    int now,next,res;
    res=heap[1];
    heap[1]=heap[len--];
    now=1;
    while(now*2<=len){
    next=now*2;
    if(next<len&&heap[next]>heap[next+1]){
    next++;
    }
    if(heap[now]<=heap[next]){
    return res;
    }
    swap(heap[now],heap[next]);
    now=next;
    }
    return res;
    }
    void work(){
    int num,lum;
    int sum=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
    scanf("%d",&num);
    put(num);
    }
    for(int i=1;i<n;i++){
    num=get();
    lum=get();
    sum+=num+lum;
    put(num+lum);
    }
    printf("%d\n",sum);
    }
    int main()
    {
    work();
    return 0;
    }

  • 0
    @ 2016-06-22 11:53:45

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    int n,a[30001],tmp,n1,ans=0;
    void shift(int r,int n)
    {int v=a[r];
    int k=2*r;
    while(k<=n)
    {
    if((a[k]>a[k+1])&&k<n)
    k++;
    if(v<a[k])
    break;
    else
    {a[r]=a[k];
    r=k;
    k=k*2;}
    }
    a[r]=v;
    }
    void jianli()
    {int r;
    for(r=n/2;r>=1;r--)
    shift(r,n);
    }
    int main()
    {
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    jianli();
    n1=n;
    for(int i=1;i<=n-1;i++)
    {
    tmp=a[1];
    a[1]=a[n1];
    n1--;
    shift(1,n1);
    a[1]=tmp+a[1];
    ans=ans+a[1];
    shift(1,n1);
    }
    cout<<ans;
    return 0;
    }

  • 0
    @ 2016-06-12 10:25:34

    不是一个优先队列就没了吗= =时间也很优越啊。
    难道你们都哪来练习哈弗曼树吗= =

  • 0
    @ 2016-04-06 16:21:27

    测试数据 #0: Accepted, time = 218 ms, mem = 880 KiB, score = 10

    测试数据 #1: Accepted, time = 15 ms, mem = 880 KiB, score = 10

    测试数据 #2: Accepted, time = 0 ms, mem = 880 KiB, score = 10

    测试数据 #3: Accepted, time = 0 ms, mem = 884 KiB, score = 10

    测试数据 #4: Accepted, time = 31 ms, mem = 880 KiB, score = 10

    测试数据 #5: Accepted, time = 46 ms, mem = 880 KiB, score = 10

    测试数据 #6: Accepted, time = 187 ms, mem = 880 KiB, score = 10

    测试数据 #7: Accepted, time = 218 ms, mem = 884 KiB, score = 10

    测试数据 #8: Accepted, time = 187 ms, mem = 880 KiB, score = 10

    测试数据 #9: Accepted, time = 234 ms, mem = 880 KiB, score = 10

    Accepted, time = 1136 ms, mem = 884 KiB, score = 100

    program p1097;
    var
    a:array[1..20000] of longint;
    i,n,j,y,m,ans,x:longint;
    procedure qsort(l,r:longint);
    var i,j,mid,p:longint;
    begin
    i:=l;j:=r;
    mid:=a[(l+r) div 2];
    repeat
    while a[i]<mid do
    i:=i+1;
    while a[j]>mid do
    j:=j-1;
    if i<=j then
    begin
    p:=a[i];a[i]:=a[j];a[j]:=p;
    i:=i+1;
    j:=j-1;
    end;
    until i>j;
    if l<j then qsort(l,j);
    if i<r then qsort(i,r);
    end;
    begin
    ans:=0;
    readln(n);
    for i:=1 to n do
    read(a[i]);
    qsort(1,n);
    for i:=1 to n-1 do
    begin
    x:=a[i]+a[i+1];
    ans:=ans+x;
    a[i+1]:=x;
    for j:=i+1 to n-1 do
    begin
    if a[j]>a[j+1] then
    begin
    m:=a[j];
    a[j]:=a[j+1];
    a[j+1]:=m;
    end;
    end;
    end;
    write(ans);
    end.

  • 0
    @ 2016-03-31 12:00:56

    直接优先队列就行了
    #include <cstdio>
    #include <iostream>
    #include <queue>
    using namespace std;
    priority_queue <int,vector <int>,greater <int> >p;
    int n,ans=0;
    int main(){
    scanf ("%d",&n);
    for (int i=1;i<=n;i++){
    int x;
    scanf ("%d",&x);
    p.push(x);
    }
    while (p.size()>1){
    int a=p.top();
    p.pop();
    int b=p.top();
    p.pop();
    ans+=a+b;
    p.push(a+b);
    }
    printf ("%d\n",ans);
    return 0;
    }

  • 0
    @ 2016-03-22 13:23:06

    手写了个堆排。。稍微有点菜啊。。

    #include <iostream>
    
    using std::cin;
    using std::cout;
    using std::endl;
    
    class Heap{ //手写排序用堆_(:з」∠)_ 
        private:
            int * data;
            int head;
            int len;
            int total;
            int hsize;
        protected:
            int left_child(int x){
                return (x-head+1)*2+head-1;
            }
            int right_child(int x){
                return (x-head+1)*2+head;
            }
            int parent(int x){
                return (x-head+1)/2+head-1;
            }
            void fix_sub(int x){
                if(x>=len)
                    return;
                int smallest = x;
                if(left_child(x)<len&&data[left_child(x)]<data[smallest]){
                    smallest = left_child(x);
                }
                if(right_child(x)<len&&data[right_child(x)]<data[smallest]){
                    smallest = right_child(x);
                }
                if(smallest!=x){
                    int swp = data[x];
                    data[x] = data[smallest];
                    data[smallest] = swp;
                    fix_sub(smallest);
                }
            }
        public:
            Heap(int * p,int length){
                this->data = p;
                this->head = 0;
                this->total = 0;
                this->len = length;
                this->hsize = length;
            }
            void build_heap(){
                for(int x=(len+head)/2;x>=head;x--){
                    fix_sub(x);
                }
            }
            int size(){
                return hsize;
            }
            int pop(){
                head++;
                hsize--;
                int tmp = data[len-1];
                data[len-1] = data[head];
                data[head] = tmp;
                build_heap();
                return data[head-1];
            }
            void insert(int x){ //这个方法很不安全。。仅仅对于这道题适用 
                hsize++;
                head--;
                data[head]=x;
                fix_sub(head);
            }
            int next(){
                int a=pop(),b=pop();
                total+=a+b;
                insert(a+b);
                return size();
            }
            void operator --(int x){
                total+=data[len-1]+data[len-2];
            }
            int operator ++(int x){
                return next();
            }
            operator int(){
                return total;
            }
            void to_string(){
                for(int x=head;x!=head+hsize;x++){
                    cout<<x<<"-"<<data[x]<<" ";
                }
                cout<<endl;
            }
    };
    
    int main(int argc,char ** argv){
        int n,*a;
        cin>>n;
        a = new int[n];
        for(int x=0;x!=n;x++){
            cin>>a[x];
        }
        Heap* heap = new Heap(a,n); //堆排序?2333 
        heap->build_heap(); //建堆 
        /*for(int x=0;x!=n;x++){
            cout<<a[x]<<endl;
        }*/
        while(((*heap)++)>2); //论运算符重载_(:з」∠)_ 
        (*heap)--;
        cout<<(int)(*heap)<<endl;
        return 0;
    }
    
  • 0
    @ 2016-03-12 16:05:33

    这题弱智

    • @ 2016-03-12 16:05:54

      #include <vector>
      #include <algorithm>
      #include <iostream>
      #include <queue>
      using namespace std;
      int main()
      {
      int n,k,i,a,b;
      cin>>n;
      priority_queue<int, vector<int>, greater<int> >P;
      for(i=0;i<n;i++)
      {
      cin>>k;
      P.push(k);
      }
      int he=0;
      for(i=1;i<n;i++)
      {
      a=P.top();
      P.pop();
      b=P.top();
      P.pop();
      he+=a+b;
      P.push(a+b);
      }
      cout<<he<<endl;
      system ("pause");
      return 0;
      }

    • @ 2016-03-12 16:06:15

      niruozhi

  • 0
    @ 2016-03-12 16:02:00

    #include <queue>
    using namespace std;
    int main()
    {
    int n,i,a,b;
    cin>>n;
    priority_queue<int, vector<int>, greater<int> > Q;
    for(i=0;i<n;i++)
    {
    cin>>a;

    Q.push(a);

    }
    int ans=0;
    for(i=1;i<n;i++)
    {
    a=Q.top();
    Q.pop();
    b=Q.top();
    Q.pop();
    ans+=a+b;
    Q.push(a+b);

    }
    cout<<ans<<endl;

    //system("pause") ;
    return 0;
    }

  • 0
    @ 2016-03-12 16:01:24

    编译成功

    测试数据 #0: Accepted, time = 15 ms, mem = 704 KiB, score = 10

    测试数据 #1: Accepted, time = 0 ms, mem = 552 KiB, score = 10

    测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10

    测试数据 #4: Accepted, time = 15 ms, mem = 592 KiB, score = 10

    测试数据 #5: Accepted, time = 0 ms, mem = 636 KiB, score = 10

    测试数据 #6: Accepted, time = 15 ms, mem = 704 KiB, score = 10

    测试数据 #7: Accepted, time = 15 ms, mem = 704 KiB, score = 10

    测试数据 #8: Accepted, time = 15 ms, mem = 704 KiB, score = 10

    测试数据 #9: Accepted, time = 15 ms, mem = 700 KiB, score = 10

    Accepted, time = 90 ms, mem = 704 KiB, score = 100

信息

ID
1097
难度
6
分类
贪心 点击显示
标签
递交数
23906
已通过
6330
通过率
26%
被复制
41
上传者