HDU 2602 Bone Collector WA谁来帮忙找找错

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big
bag with a volume of V ,and along his trip of collecting there are a lot of
bones , obviously , different bone has different value and different volume, now
given the each bone’s value along his trip , can you calculate out the maximum
of the total value the bone collector can get ?

 

Input

The first line contain a integer T , the number of
cases.
Followed by T cases , each case three lines , the first line contain
two integer N , V, (N <= 1000 , V <= 1000 )representing the number of
bones and the volume of his bag. And the second line contain N integers
representing the value of each bone. The third line contain N integers
representing the volume of each bone.

 

Output

One integer per line representing the maximum of the
total value (this number will be less than 2³¹).

 

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

 

Sample Output

14

 

解题心得：

　　这是一个背包问题，但是还没学会动态规划的方法，只好用贪心法求解，但提交一直是wronganswer，谁看到了请告诉我。

 

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef struct{
    int val;
    int vol;
    float vv;
}Bone;

bool compare(Bone a,Bone b)
{
    if(a.vv==b.vv){
        return a.vol>b.vol;
    }
    return a.vv>b.vv;
}

int main()
{
    int t;//t组测试数据
    int n;//n个骨头
    int v;//书包能装的体积
    int now_v=;//已装入的体积
    int j1=;//已装入的个数
    int sum=;//已装入的总价值
    Bone bone[];
    cin>>t;
    for(int i=;i<t;i++){
        scanf("%d %d",&n,&v);
        for(int j=;j<n;j++){
            scanf("%d",&bone[j].val);
        }
        for(int j=;j<n;j++){
            scanf("%d",&bone[j].vol);
        }
        for(int j=;j<n;j++){
            bone[j].vv=(float)bone[j].val/(float)bone[j].vol;
        }
        sort(bone,bone+n,compare);
        //for(int j=0;j<n;j++){
        //    cout<<bone[j].vv<<" ";
        //    cout<<bone[j].val<<" "<<endl;
        //}
        for(int j=;j<n;j++){
            now_v+=bone[j].vol;//循环一次往里装一次
            j1++;
            if(now_v>=v){
                break;
            }
        }
        if(now_v==v){
            for(int j=;j<j1;j++){
                sum+=bone[j].val;
            }
        }
        if(now_v>v){
            for(int j=;j<j1-;j++){
                sum+=bone[j].val;
            }
        }
        if(now_v<v){
            for(int j=;j<n;j++){
                sum+=bone[j].val;
            }
        }
        cout<<sum<<endl;
        now_v=;
        sum=;
        j1=;
    }
    return ;
}