问题描述:
给定n种物品和一背包。物品i的重量是wi,其价值为vi,背包容量为c。问应如何选择装入背包中的物品,使得装入背包中物品的总价值最大。在选择装入背包的物品时,对每种物品i只有两种选择,即装入背包或不装入背包。不能将物品i装入背包多次,也不能只装入部分的物品i。因此,该问题称为0-1背包问题。
DP思想:
首先物品个数为n个, 包总容量为c, 其中w[],v[]分别为n个物品的重量、价格。表m(i,j)表示从第i个物品开始容量为j的包的最大价值量。具体流程如下:
1. 初始化,当i=n-1时,若c>=w[i],则m(i,c)=v[i];若0<=j<w[i],则m(i,c)=0
2. 当0<=i<n-1时,若j>=w[i],则m(i,j)=max{ m(i+1,j), m(i+1,j-w[i])+v[i] };若0<=j<w[i],则m(i,j)=m(i+1,j)
C++源码:
class knap{
public:
knap();
void init();
int run();
void dumpW();
void dumpV();
void dumpTable();
void dumpTrace();
int getCap();
int getNum();
private:
void trace();
const static int NUM_OF_GOODS = 256;
const static int CAPACITY = 256;
int w[NUM_OF_GOODS];//weight
int v[NUM_OF_GOODS];//value
int c;//capacity
int num;//num of goods
int table[NUM_OF_GOODS][CAPACITY];
bool isIn[NUM_OF_GOODS];
};
knap::knap(){
memset(w, 0, sizeof(w));
memset(v, 0, sizeof(v));
memset(table, 0, sizeof(table));
memset(isIn, 0, sizeof(isIn));
num = c = 0;
}
void knap::init(){
num = 5;
c = 10;
w[0] = 1 , w[1] = 3, w[2] = 2, w[3] = 9, w[4] = 4;
v[0] = 10, v[1] = 5, v[2] = 6, v[3] = 8, v[4] = 7;
}
int knap::run(){
for(int i=0;i<=c;i++){
if(w[num-1]<=i)
table[num-1][i] = v[num-1];
else
table[num-1][i] = 0;
}
for(int i=num-2;i>=0;i--){
for(int j=0;j<=c;j++){
if(j>=w[i]){
table[i][j]=max(table[i+1][j], table[i+1][j-w[i]]+v[i]);
}else{
table[i][j]=table[i+1][j];
}
}
}
this->trace();
return table[0][c];
}
void knap::trace(){
int curCap = c;
for(int i=0;i<num-1;i++){
if(table[i][curCap]==table[i+1][c])
isIn[i] = false;
else{
isIn[i] = true;
curCap -= w[i];
}
}
isIn[num-1] = curCap >= w[num-1] ? true : false;
}
void knap::dumpW(){
for(int i=0;i<num;i++){
cout << w[i] << "/t";
}
cout << endl;
}
void knap::dumpV(){
for(int i=0;i<num;i++){
cout << v[i] << "/t";
}
cout << endl;
}
void knap::dumpTable(){
for(int i=num-1;i>=0;i--){
for(int j=0;j<=c;j++)
cout << table[i][j] << "/t";
cout << endl;
}
}
void knap::dumpTrace(){
for(int i=0;i<num;i++){
cout << isIn[i] << "/t";
}
cout << endl;
}
int knap::getCap(){
return c;
}
int knap::getNum(){
return num;
}
int main()
{
knap k;
int ans;
k.init();
ans = k.run();
cout << "Num of goods:/t" << k.getNum() << endl;
cout << "Capacity:/t" << k.getCap() << endl;
cout << "Weight:/n";
k.dumpW();
cout << "Value:/n";
k.dumpV();
cout << "Table:/n";
k.dumpTable();
cout << "Trace:/n";
k.dumpTrace();
cout << "answer:/n" << ans << endl;
return 0;
}
运行结果: