0-1背包问题之分支限界法

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有n个物体，重量和价值已知，要放入容量为c的背包里，放入的时间，要求放入的总重量不能超过包的容量，同时保证价值最大。

前面分别用回溯法、暴力搜索法进行了求解，这里利用分支限界法进行求解。参考代码如下：

#include <stdio.h>#include <stdlib.h>typedef struct QNode{int value;//当前结点的总价值int weight; //当前的总重量struct QNode *next;}QNode, *QueuePtr;typedef struct{QueuePtr front;QueuePtr rear;}Queue;int initQueue(Queue &Q){Q.front=Q.rear=(QueuePtr)malloc(sizeof(QNode));if(!Q.front)return -1;Q.front->next=NULL;return 1;}int emptyQueue(Queue Q){if (Q.front==Q.rear)return 1;elsereturn 0;}int destroyQueue(Queue &Q){while(Q.front){Q.rear=Q.front->next;free(Q.front);Q.front=Q.rear;}return 1;}int enQueue(Queue &Q, int value, int weight){QueuePtr p=(QueuePtr)malloc(sizeof(QNode));if(!p)return -1;p->value=value; p->weight=weight; p->next=NULL;Q.rear->next=p;Q.rear=p;return 1;}int deQueue(Queue &Q, int &value, int &weight){QueuePtr p;if(Q.front==Q.rear)return -1;p=Q.front->next;value=p->value;weight=p->weight;Q.front->next=p->next;if(Q.rear==p) Q.rear=Q.front;free(p);return 1;}Queue loadingQueue;int bestvalue, n;void inQueue(int value, int weight, int i){if(i==n-1){if(value>bestvalue)bestvalue=value;}elseenQueue(loadingQueue,value,weight);}int main(){int i,j,k;int *w, *v, ew, ev;int c;printf("input the number of things and the volume of ships:");scanf("%d%d",&n,&c);w=new int[n];v=new int[n];printf("input the weights:");for(i=0;i<n;i++)scanf("%d",&w[i]);printf("input the values:");for(i=0;i<n;i++)scanf("%d",&v[i]);initQueue(loadingQueue);enQueue(loadingQueue,-1,0);//int deQueue(Queue &Q, int &value, int &weight)//void inQueue(int value, int weight, int i)//int enQueue(Queue &Q, int value, int weight)i=0;//层数ew=0;//扩展结点对应的载重量ev=0;while(true){if(ew+w[i]<=c)inQueue(ev+v[i],ew+w[i],i);inQueue(ev,ew,i);deQueue(loadingQueue, ev,ew);if(ev==-1) //同层结点尾部{if(emptyQueue(loadingQueue)){printf("the result is %d.\n",bestvalue);}enQueue(loadingQueue,-1,0);deQueue(loadingQueue, ev,ew);i++;}}return 0;}

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