回溯法 0 1背包问题

用回溯法解决0 1背包问题

#include <stdio.h>
#include <stdlib.h>

/* 用回溯法解决01背包问题,维护的变量有： curState[]: 当前状态中物品是否被选中，若i选中,则curState[i]=1,否则为0. optState[]: 从搜索开始到目前状态中的最优状态。 curWeight: 当前的总重量 curValue: 当前的总价值 maxValue: 最佳状态的价值 curDeep: 当前深度 N: 总的物品个数 capacity: 背包的容量 weight[]: 第i个物品的重量为weight[i] value[]: 第i个物品的价值value[i] */

int N,capacity;
int *weight,*value;
int maxValue = 0;
int *curState,*optState;

void backtracking(int curDeep,int curWeight,int curValue){

    int i;

    if(curDeep>=N){     //到达叶节点
        if(curWeight<=capacity && curValue>maxValue){

            maxValue = curValue;        //更新最大值和最大状态向量
            for(i=0;i<N;i++){
                optState[i] = curState[i];
            }
        }

    }else{
        curState[curDeep] = 1;  //选中
        if(curWeight+weight[curDeep]<=capacity)
            backtracking(curDeep+1,curWeight+weight[curDeep],curValue+value[curDeep]);

        curState[curDeep] = 0;  //不选
        backtracking(curDeep+1,curWeight,curValue);
    }
}

int main(){

    int i,tw,tv;

    printf("input number of object:\n");
    scanf("%d",&N);
    printf("input capacity of bag:\n");
    scanf("%d",&capacity);

    weight = (int *)malloc(sizeof(int)*N);
    value = (int *)malloc(sizeof(int)*N);
    curState = (int *)malloc(sizeof(int)*N);
    optState = (int *)malloc(sizeof(int)*N);

    printf("input weight and value,by format: weight value\n");
    for(i=0;i<N;i++){
        scanf("%d %d",&tw,&tv);
        weight[i] = tw;
        value[i] = tv;
    }

    backtracking(0,0,0);
    printf("%d\n",maxValue);

    return EXIT_SUCCESS;
}