CODEVS 2144 砝碼稱重2
阿新 • • 發佈:2019-02-04
//二分dfs,最後合併
#include<bits/stdc++.h>
using namespace std;
const int maxn = 400000;
int n, m, tot1, tot2, a[31];
int ans = 0x3f3f3f3f;
struct node{
int num, w;
} hash1[maxn], hash2[maxn];
bool cmp(node p, node q){
return p.w < q.w;
}
void dfs1(int haoyu, int step, int mnow){
if(haoyu > n/2 ) return;
hash1[++tot1].num = step + 1;
hash1[tot1].w = mnow + a[haoyu];
//if(hash1[tot1].w == m) ans = hash1[tot1].num;
dfs1(haoyu+1, step+1, mnow+a[haoyu]);
dfs1(haoyu+1, step, mnow);
}
void dfs2(int haoyu, int step, int mnow){
if(haoyu > n) return;
hash2[++tot2].num = step + 1;
hash2[tot2].w = mnow + a[haoyu];
//if(hash2[tot2].w == m) ans = hash2[tot2].num;
dfs2(haoyu+1, step+1, mnow+a[haoyu]);
dfs2(haoyu+1, step, mnow);
}
int find(int x){//二分查詢程式碼要記牢,左開右閉
int l = 1, r = tot2+1, mid = l + (r-l)/2;
while(l < r){
if(x < hash2[mid].w){ r = mid; mid = l + (r-l)/2 ; }
if(x > hash2[mid].w){ l = mid + 1; mid = l + (r-l)/2; }
if(x == hash2[mid].w) return mid;
}
return -1;
}
int main(){
cin >> n >> m;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a+1, a+n+1);
dfs1(1, 0, 0);
sort(hash1+1, hash1+tot1+1, cmp);
dfs2(n/2+1, 0, 0);
sort(hash2+1, hash2+tot2+1, cmp);
for(int i = 0; i <= tot1; i++){//從0開始是因為有可能前半段沒有目標砝碼,全在後半段,如果從1開始就找不到後半段的了
int flag = find(m - hash1[i].w);
if(flag != -1){
ans = min(hash1[i].num + hash2[flag].num, ans);
}
}
cout << ans << endl;
return 0;
}