結構struct動態數組創建、操作、刪除

轉載地址：> http://blog.csdn.net/lawrencesgj/article/details/7939755

在做hdoj的1009，本來這道題目不是很難，可是對於struct動態數組操作不是很熟，做了很久，在這裏記錄一下，避免下次繼續出錯。
FatMouse’ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25384 Accepted Submission(s): 8029

Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1’s. All integers are not greater than 1000.

Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output
13.333
31.500

代碼如下：

#include<stdio.h>
#include<algorithm>
using namespace std;

struct trade{
    int j;
    int f;
};

bool compare(const trade &a, const trade &b){
    return double(a.j)/double(a.f)>double(b.j)/double(b.f);
}

void main(){
    int m,n;
    double sum;
    while(1){
        sum=0.0;
        scanf("%d %d",&m,&n);
        if(m==-1)
            break;
        trade* array=new trade[n];
        for(int i=0; i<n; i++){
            scanf("%d %d",&array[i].j,&array[i].f);
        }
        sort(array,array+n,compare);
        for(int j=0; j<n; j++){
            if(m<array[j].f){
                sum+=double(array[j].j)/double(array[j].f)*m;
                break;
            }
            m-=array[j].f;
            sum+=array[j].j;
        }
        printf("%.3f\n",sum);
        delete []array;
    }
}

其中，創建動態struct數組

struct struct_name * struct_array = new struct struct_name[struct_array_len];
//or discard struct 
struct_name * struct_array = new struct_name[struct_array_len];  

輸入struct數組元素值，記得不要落下&

scanf("%d",&struct_array[i].struct_member);

排序

sort(struct_array,struct_array+struct_array_len,compare);

釋放

delete []struct_array;