一 问题描述：有一批共个集装箱要装上2艘载重量分别为C1和C2的轮船，其中集装箱i的重量为Wi，且
容易证明：如果一个给定装载问题有解，则采用下面的策略可得到最优装载方案。
(1)首先将第一艘轮船尽可能装满；
(2)将剩余的集装箱装上第二艘轮船。
二：
（1）解决装载问题的优先队列，将扩展节点的当前装载的重量(Ew)加上剩余集装箱的总重量 (r[i]) 作为优先级，从优先级队列中出队列。
（2）为了方便找到最优的分配方法，在每一个节点设置指向父节点的指针，当算法走到第 n+1层就可以停止寻找，因为当前扩展节点队列 而 i = n+1 时，说明了当前的优先级 uweight最高，就是说 Ew +r[n+1]最大，而r[n+1] =0 , 所以Ew要大于队列中剩余的所有的Ew +r[i] ;
#include <iostream>
#include <algorithm>
using namespace std;
template <typename T>
class MaxHeap{
public:
MaxHeap(int cap);
bool insert(T value);
bool remove(T value);
void print();
bool createMaxHeap(T *a,int length);
T getTop();
void filterup(int index);
void filterdown(int start,int end);
int size;
private:
int capacity;
T *heap;
};
template <typename T>
MaxHeap<T>::MaxHeap(int cap):capacity(cap),size(0),heap(NULL){
heap = new T[capacity];
};
template <typename T>
T MaxHeap<T>::getTop(){
if(size != 0){
size--;
T temp = heap[0];
heap[0] = heap[size];
filterdown(0,size);
return temp;
}
}
template <typename T>
void MaxHeap<T>::print(){
for(int i = 0; i < size;i++){
cout<<heap[i];
if(i == size-1)
cout<<endl;
else
cout<<" ";
}
}
template <typename T>
bool MaxHeap<T>::createMaxHeap(T *a,int length){
if(length > capacity)
return false;
for(int i = 0; i < length;i++)
heap[i] = a[i];
size = length;
int currentPos = (size-2)/2;
while(currentPos >= 0){
filterdown(currentPos,size-1);
currentPos--;
}
return true;
}
template <typename T>
void MaxHeap<T>::filterup(int start){
int j = start;
if(j == 0)
return;
int i = (start-1)/2;
T temp = heap[start];
while( j > 0){
if(heap[i]<= temp){
heap[j] = heap[i];
j = i;
i = (i-1)/2;
}else
break;
}
heap[j] = temp;
}
template<typename T>
void MaxHeap<T>::filterdown(int start,int end){
int i = start;
int j = i*2 + 1;
T temp = heap[i];
if(j > size-1)
return;
while( i <= end){
if(j > size-1)
break;
if(j + 1 <=size-1 && heap[j] < heap[j+1])
j++;
if(temp > heap[j])
break;
else{
heap[i] = heap[j];
i = j;
j = 2*i+1;
}
}
heap[i] = temp;
}
template <typename T>
bool MaxHeap<T>::remove(T value){ //删除节点
bool judge = false;
for(int i = 0; i < size;i++){
if(heap[i] == value){
heap[i] = heap[size-1];
size--;
filterdown(i,size);
judge = true;
}
}
return judge;
}
template <typename T>
bool MaxHeap<T>::insert(T value){
if(size == capacity){
cout<<"Heap full"<<endl;
return false;
}
heap[size] = value;
filterup(size);
size++;
return true;
}
template <typename Type>class HeapNode;
class TNode{
public:
friend bool AddactiveNode(MaxHeap<HeapNode<int> >&,TNode*,int ,bool ,int );
friend int Maxloading(int w[],int ,int ,bool *);
private:
TNode *father;
bool lchild;
};
template <typename Type>class HeapNode{
public:
friend bool AddactiveNode(MaxHeap<HeapNode<int> >&,TNode*,int ,bool ,int );
friend int Maxloading(int w[],int,int ,bool *);
operator Type()const{
return uweight;}
friend bool operator <=(const HeapNode &a,const HeapNode &b){
return a.uweight < b.uweight;
}
private:
TNode *ptr;
Type uweight;
int level;
};
bool AddactiveNode(MaxHeap<HeapNode<int> >&H,TNode *E,int wt,bool ch,int lev){//向队列中插入一个节点
TNode *node = new TNode;
node->lchild = ch;
node->father = E;
HeapNode<int>op;
op.ptr = node;
op.uweight = wt;
op.level = lev;
H.insert(op);
return true;
}
int Maxloading(int w[],int c,int n,bool bestx[]){
int *r;
r = new int[n+1];
r[n] = 0; //数组r[i]用来记录第i个集装箱之后的总重量 （不包括第i个集装箱的重量）
for(int j = n-1; j >= 0;j--)
r[j] = r[j+1] + w[j+1];
MaxHeap<HeapNode<int> >H(100000);
int i =1; //扩展节点初始为第1层
int bestw = 0;
TNode *E = 0; //指向扩展节点的指针
int Ew = 0; //扩展节点的当前装载为Ew
while(i!=n+1){ //到达第n+1层就停止
if(Ew + w[i] <= c)
//添加左孩子
AddactiveNode(H,E,Ew+w[i]+r[i],true,i+1); //左孩子标记true
//添加右孩子
AddactiveNode(H,E,Ew+r[i],false,i+1); //右孩子标记false
HeapNode<int>T;
//更新扩展节点
T = H.getTop(); //得到优先级最高的节点，并删除这个节点
i = T.level;
E = T.ptr;
Ew = T.uweight - r[i-1];
}
for(int j = n;j >0;j--){ //根据左孩子标记判断是否装入
bestx[j] = E->lchild;
E = E->father;
}
return Ew;
}
int main(){
int w[7] = { 0,1,2,3,4,5,6};
int aw = 0;
for(int i = 0;i < 7;i++)
aw +=w[i];
int c1 = 19;
int c2 = 20;
bool bestxp[7];
int w1 = Maxloading(w,c1,6,bestxp);
if(c2 < aw - w1)
return false;
for(int i = 1;i < 7;i++){
if(bestxp[i] == true)
cout<<"第"<<i<<"件物品"<<"装入轮船1"<<endl;
else if(bestxp[i] == false)
cout<<"第"<<i<<"件物品"<<"装入轮船2"<<endl;
}
return 0;
}