Greedy method--boat problem

The topics are generally:
There are n individuals, the personal weight of w[i], the maximum load per ship is C, and can only be up to two people. Load the owner with the fewest ships.

Solution:
First of all, from the lightest, he should sit with the heaviest man, and if everyone could not ship with it, then the only way is for everyone to be a boat.
Otherwise, he should choose the heaviest person who can take a boat with him, and this will not be wasted.

#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream>using namespace Std;struct Node{int W;} A[1001];bool CMP (node A,node b) {return A.W<B.W;} int main () {int i,j,n,c;int book[1001]={0};scanf ("%d%d", &n,&c), for (i=1;i<=n;i++) scanf ("%d", &A[I].W) ; sort (a+1,a+1+n,cmp); int num=0;for (i=1;i<=n;i++) {//book[i]=1; for (j=n;j>=1;j--) {//one: If they are two weight and less than or equal to C, And J This man has not been pulled away by others, then this is a kind; if (a[i].w+a[j].w<=c && book[j]==0) {num++; book[i]=1; book[j]=1; break;} Two: If they weigh two and greater than C and J is not pulled, then J will only be able to sit alone with himself in a boat; else if (a[i].w+a[j].w>c && book[j]==0) {num++; Book[j]=1; }}//If the above are all circulating, but I still do not meet the conditions, then it is also only to sit on their own boat, if (!book[i])  {num++; book[i]=1;}} printf ("%d\n", num);}

Greedy method--boat problem