c++ STL平常练习-6

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struct tank
{
    double dis;
    double price;

    bool operator < (const tank &t)
    { return dis < t.dis; }
}tanks[101];

int main117()               //贪心解决加油站问题，到达目的花费最少
{
    double tankCap,distance,disUnit;
    int n,i,j;
    bool flag = true;

    while(cin>>tankCap>>distance>>disUnit>>n)
    {
        flag = true;
        for(i=0; i<n; ++i)
            cin>>tanks[i].price>>tanks[i].dis;
        tanks[n].price = 0; tanks[n].dis = distance;
        sort(tanks,tanks+n);

//for(i=0; i<=n; ++i)
//  cout<<tanks[i].price<<"  "<<tanks[i].dis<<endl;

        double curOil = 0, oilMoney = 0;
        double Dis = disUnit * tankCap; //满油的状态能跑多远
        for(i=0; i<n; )                                                  //里面包含三个循环进行查找，其中前两个循环都是找比当前站油价低的站，最后一个找当前站后面的最低油价的站
        {// cout<<i<<endl;
            if(tanks[i+1].dis - tanks[i].dis > Dis)
            { flag = false; cout<<"The maximum travel distance is: "<<tanks[i].dis+Dis<<endl; break; }

            double min_price = tanks[i].price;
            int min_pos = i;
            for(j=i+1; j<=n && tanks[j].dis <= tanks[i].dis + curOil * disUnit; ++j)  //先查找当前的油量之内的距离内的最小油价,必须是当前油量之内的
            {
                if(min_price > tanks[j].price)
                {
                    min_price = tanks[j].price;
                    min_pos = j;
                }
            }

            if(i != min_pos)   //表示找到
            {
                curOil -= (tanks[min_pos].dis-tanks[i].dis)/disUnit;
                i = min_pos;
    //          cout<<"one one: "<<curOil<<"   "<<i<<endl; system("pause");
                continue;
            }

            min_price = tanks[i].price;
            min_pos = i;
            for(j=1+i; j<=n && tanks[j].dis <= tanks[i].dis + Dis; ++j)   //此步查找距离当前站最近的并且油价比当前站还低
            {
                if(min_price > tanks[j].price)
                {
                    min_pos = j;
                    break;
                }
            }
            if(min_pos != i)   //表示找到
            {
                double oilGap = (tanks[min_pos].dis-tanks[i].dis)/disUnit - curOil; //表示行驶到min_pos点处，需要再加多少油
                curOil = 0;
                oilMoney += (oilGap*tanks[i].price);
                i = min_pos;
    //          cout<<"two two: "<<curOil<<"   "<<i<<endl; system("pause");
                continue;
            }

            min_price = 0XFFFFFFF;
            min_pos = i;
            for(j=1+i; j<=n && tanks[j].dis <= Dis +tanks[i].dis; ++j) //此处查找满油状态下不算本站，油价最低的站并且一定可以找到
            {
                if(min_price > tanks[j].price)
                {
                    min_price = tanks[j].price;
                    min_pos = j;
                }
            }
            oilMoney += (tankCap-curOil)*tanks[i].price;
            curOil  = tankCap - (tanks[min_pos].dis-tanks[i].dis)/disUnit;
            i = min_pos;
//          cout<<"three three: "<<curOil<<"   "<<i<<endl; system("pause");
        }
        if(flag) cout<<fixed<<setprecision(2)<<oilMoney<<endl;
    }

    return 0;
}

string getPre(int n)
{
    int iarr[16];
    int index = 0;
    int i = 0;
    while(n)
    {
        if((0X00000001 & n) != 0) //表示此处是1
            iarr[index++] = i;
        ++i;
        n>>=1;
    //  cout<<n<<endl;
    }
    string s ;
    for(i=index-1; i>=0; --i)
    {
        if(iarr[i] == 2)
        {
            if(i == 0) s += "2(2)";
            else s += "2(2)+";
    //      cout<<iarr[i]<<"   "<<s<<endl;
    //      system("pause");
        }else if(iarr[i] == 1)
        {
            if(i == 0) s += "2";
            else s += "2+";
    //      cout<<iarr[i]<<"   "<<s<<endl;
    //      system("pause");
        }else if(iarr[i] == 0)
        {
            if(i == 0) s += "2(0)";
            else s += "2(0)+";
    //      cout<<iarr[i]<<"   "<<s<<endl;
    //      system("pause");
        }
        else {
            if( i == 0) s += ("2("+getPre(iarr[i])+")");
            else s += ("2("+getPre(iarr[i])+")+");
    //      cout<<iarr[i]<<"   "<<s<<endl;
    //      system("pause");
        }
    }
    return s ;
}


int main119()
{
    int n;
    while(cin>>n)
    {
        string s = getPre(n);
        cout<<s <<endl;
    }
    return 0;
}
/*
int main120()
{
//  char tree[26][26];
    vector<string> tree[26];
    char str[80];
    bool root[26];
    memset(root,true,sizeof(root));
    while(gets(str))
    {
        int i=0;
        char parent = '*';
        while(str[i] != 0)
        {
            if(isalpha(str[i]))
            {
                if(parent == '*')
                    parent = str[i];
                else
                {
                    tree[parent-'a'].push_back(str[i]);
                    parent = str[i];
                    root[str[i]-'a'] = false;
                }
            }
            ++i;
        }
    }
    return 0;
}
*/
void deleteNode(BTree3 *&root,int key)
{
    BTree3 *f = NULL, *p = root;

    while(p!=NULL && p->data!=key)
    {
        f = p;
        if(p->data > key)
            p = p->lchild;
        else p = p->rchild;
    }

    if(p!=NULL)  //表示在排序二叉树中找到了值为key的节点，若找不到则一定是到达了NULL节点
    {               //如果f为空则表示找到的是根节点
            if(key == 4)
            {
                if(p->lchild != NULL) cout<<"left : "<<p->lchild->data;
                if(p->rchild != NULL) cout<<"right : "<<p->rchild->data<<endl;
            }

        if(p->lchild == NULL && p->rchild == NULL)     //节点是叶节点
        {
            if(f)
            {
                if(f->lchild == p)
                    f->lchild = NULL;
                else f->rchild = NULL;
                free(p);
                p = NULL;
            }else
            {
                free(p);
                root = NULL;
            }
        }else if(p->lchild == NULL)          //节点的左节点为空
        {
            if(key == 4)
                cout<<"左节点为空"<<endl;
            if(f)
            {
                if(f->lchild == p)
                    f->lchild = p->rchild;
                else if(f->rchild == p)
                    f->rchild = p->rchild;
                free(p);
            }else
            {
                BTree3 *tmp = p;
                p = p->rchild;
                root = p;
                free(tmp);
            }
        }else if(p->rchild == NULL)     //节点的右节点为空
        {
            if(f)
            {
                if(f->lchild == p)
                    f->lchild = p->lchild;
                else
                    f->rchild = p->lchild;
                free(p);
            }else
            {
                BTree3 *tmp = p;
                p = p->lchild;
                root = p;
                free(tmp);
            }


        }else            //节点的左右节点都不为空,   因为这种方法是把找到的节点的值进行替换，所以不需要改变节点的左右孩子指针
        {
            BTree3 *pp = p->lchild;
            BTree3 *ff = p;

            while(pp->rchild != NULL)  //此处找其左子树的最右节点
            {
                ff = pp;
                pp = pp->rchild;
            }

            if(ff == p)  //表示p的左子树的第一个节点
            {
                p->data = pp->data;
                p->lchild = pp->lchild;
                free(pp);
            }else
            {
                p->data = pp->data;
                ff->rchild = pp->lchild;
                free(pp);
            }
        }
    }
}

void Insert(BTree3 *&root,int key)
{
    if(root == NULL)
    {
        root = (BTree3 *)malloc(sizeof(BTree3));
        root->data = key;
        root->lchild = root->rchild = NULL;
    }else if(root->data > key)
        Insert(root->lchild,key);
    else Insert(root->rchild,key);
}

void  main121()
{
    int n;
    BTree3 *root = NULL;
    while(cin>>n)
    {
        for(int i=0; i<n; ++i)
        {
            int tmp;
            cin>>tmp;
            Insert(root,tmp);
        }

        postOrder3(root);
        while(cin>>n && n>=0)
        {
            deleteNode(root,n);
            postOrder3(root);
        }
    }
}
void main123()   //A^B 二分法求幂
{
    int A,B;
    while(cin>>A>>B && A && B)
    {
        int AA = A;
        A = 1;
        while(B)
        {
            if(0X00000001 & B)  //表示最低位是1
            {
                A *= AA;
                A %= 1000;
            }
            AA *= AA;
            AA %= 10000;
            B>>=1;
        }
        cout<<A<<endl;
    }
}

void judgeAVL(BTree3 *root,bool &balance,int &h)  //判断是否为平衡二叉树
{
    if(root == NULL) { balance = true; h = 0; return; }
    if(root->lchild == NULL && root->rchild == NULL)
    { balance = true;  h = 1; return; }
    int hl=0,hr=0;
    bool bl=true,br=true;
    judgeAVL(root->lchild,bl,hl);
    judgeAVL(root->rchild,br,hr);
    h = 1+(hl>hr?hl:hr);
    if(abs(hl-hr) < 2)
        balance = bl & br;
    else balance = false;
}


int root3[101];
int findRoot3(int x)
{
    if(root3[x] == -1) return x;
    else
    {
        int tmp = findRoot3(root3[x]);
        root3[x] = tmp;
        return tmp;
    }
}

void main124()
{
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=1; i<=n; ++i)
            root3[i] = -1;
        for(i=1; i<=m; ++i)
        {
            int x,y;
            cin>>x>>y;
            x = findRoot3(x);
            y = findRoot3(y);
            if(x != y)
                root3[y] = x;
        }
        int cnt = -1;
        for(i=1; i<=n; ++i)
            if(root3[i] == -1)
                ++cnt;
        cout<<cnt<<endl;
    }
}

int edges[101][101];

void main125()
{
    int n,m;
    while(cin>>n>>m)
    {
        if(n==0 && m==0) break;
        int i,j,k;
        for(i=0; i<101; ++i)
            for(j=0; j<101; ++j)
            {
                edges[i][j] = -1;   //-1表示不可达
            }
        for(i=1; i<=m; ++i)
        {
            int a,b,c;
            cin>>a>>b>>c;
            edges[a][b] = c;
            edges[b][a] = c;
        }

        for(k=1; k<=n; ++k)
        {
            for(i=1; i<=n; ++i)
            {
                for(j=1; j<=n; ++j)
                    if(edges[i][k] == -1 || edges[k][j] == -1)
                        continue;
                    if(edges[i][j] == -1 || edges[i][j] > edges[i][k]+edges[k][j])
                        edges[i][j] = edges[i][k] + edges[k][j];
            }
        }
        cout<<edges[1][n]<<endl;
    }
}