传送门
题意：在n个物品中选出部分放入承重量为W的背包，使得放入的重量C满足 W/2 ≤ C ≤ W (其中W/2上取整)。若无可行方案输出-1，否则输出所选物品个数及每个物品重量。
思路：
- 首先判断是否有个物品重量刚好在 W/2到W之间，若有直接选该物品即可(可二分也可暴力找)。
- 反之，直接从最小开始连续选择重量在1到W/2之间的物品，若C能满足条件即可。(因为已经证明不存在刚好在W/2到W的物品，所以在1到W之间的物品间选择能实现的C除了能满足条件以外就只有小于W/2)。
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{
1, 0}, {
-1, 0}, {
0, 1}, {
0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;int t, n, w, b[N];struct node{
int id, da;
}a[N];int cmp(node x, node y){
return x.da < y.da;
}signed main()
{
IOS;cin >> t;while(t --){
cin >> n >> w;int sum = 0;for(int i = 1; i <= n; i ++){
cin >> a[i].da;sum += a[i].da;a[i].id = i;}sort(a+1, a+n+1, cmp);for(int i = 1; i <= n; i ++){
b[i] = a[i].da;
// cout << a[i].da << endl;}if(a[1].da>w || sum<w/2+bool(w%2)){
cout << -1 << endl;continue;}int x = w/2+bool(w%2);int pos = lower_bound(b+1, b+n+1, x)-b;if(1<= pos&&pos<=n&&b[pos]<=w){
cout << 1 << " " << a[pos].id << endl;continue;}sum = 0, pos = -1;for(int i = 1; i <= n; i ++){
sum += a[i].da;if(sum>=x&&sum<=w) {
pos = i; break;}}if(pos==-1){
cout << -1 << endl;continue;}cout << pos << endl;for(int i = 1; i <= pos; i ++) cout << a[i].id << " ";cout << endl;}return 0;
}