J - 砝码称重
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others)
Submit Status
Problem Description
有一个物品和一些已知质量的砝码和天平，问能不能用这些砝码称出这个物品的重量（天平两边都可以放砝码）
Input
多样例输入，样例数<=20000
对于每个样例：
第一行输入两个数n，m，表示砝码数量和重物质量,1 ≤ m ≤ 1018
第二行输入n个数a1 ,a2 , ... ,an ,1018 ≥ ai+1 ≥ 3 * ai ≥ 1
Output
Case x: y
x代表第几个样例，若能称出重物的质量，y为YES，否则y为NO
Sample Input
3 7 1 3 9
Sample Output
Case 1: YES
Hint
样例为一边盘放重物和质量为3的砝码，另一边盘放质量为1和9的砝码
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long LL;
4 const int N=1e5+10;
5 const int INF=0x3f3f3f3f;
6 int cas=1,T;
7 int n,vis[40];
8 LL a[40],b[40],m;
9 bool dfs(int x,LL sum,int v)
10 {
11 if(sum==0) return 1;
12 if(x<0 || b[x]<sum || sum<0) return 0;
13 if(!vis[x])
14 {
15 vis[x]=v;
16 if(dfs(x-1,sum-a[x],v)) return 1;
17 vis[x]=0;
18 }
19 if(dfs(x-1,sum,v)) return 1;
20 return 0;
21 }
22 bool solve(LL sum,int v)
23 {
24 if(sum<0) return 0;
25 int id=0;
26 while(a[id]<=sum) id++;
27 if(dfs(id-1,sum,v)) return 1;
28 if(!vis[id-1])
29 {
30 vis[id-1]=v;
31 if(solve(sum-a[id-1],v)) return 1;
32 vis[id-1]=0;
33 }
34 if(!vis[id])
35 {
36 vis[id]=v;
37 if(solve(a[id]-sum,v^1)) return 1;
38 vis[id]=0;
39 }
40 return 0;
41 }
42 int main()
43 {
44 // freopen("1.in","w",stdout);
45 freopen("1.in","r",stdin);
46 freopen("1.out","w",stdout);
47 // scanf("%d",&T);
48 while(scanf("%d%lld",&n,&m)==2)
49 {
50 memset(vis,0,sizeof(vis));
51 a[0]=b[0]=0;
52 for(int i=1;i<=n;i++)
53 {
54 scanf("%lld",a+i);
55 b[i]=b[i-1]+a[i];
56 }
57 a[n+1]=1LL*INF*INF;
58 b[n+1]=b[n]+a[n+1];
59 vis[0]=vis[n+1]=1;
60 printf("Case %d: ",cas++);
61 if(solve(m,2))
62 {
63 int sum1=0,sum2=0;
64 puts("YES");
65 /*for(int i=1;i<=n;i++) if(vis[i]==2) printf("%d ",a[i]),sum1+=a[i];
66 printf("\n");
67 for(int i=1;i<=n;i++) if(vis[i]==3) printf("%d ",a[i]),sum2+=a[i];
68 printf("\n");
69 printf("sum1:%d sum2:%d m:%d dif:%d\n",sum1,sum2,w,sum1-sum2);*/
70 }
71 else puts("NO");
72 }
73 // printf("%d %d\n",clock(),CLOCKS_PER_SEC);
74 // printf("time=%.3f\n",(double)clock()/CLOCKS_PER_SEC);
75 return 0;
76 }
solve.cpp
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef unsigned long long ULL;
4 typedef long long LL;
5 const int INF = 0x3f3f3f3f;
6 const double eps = 1e-9;
7 long long a[40];
8 int vis[40];
9 int cas=1;
10 int check(int n, LL x, LL p)
11 {
12 for(int i=n-1;i>=0;--i)
13 {
14 if(x - a[i] >= p) x -= a[i], vis[i] = 0;
15 if(x >= p + a[i]) p += a[i], vis[i] = 2;
16 }
17 if(x == p) return 1;
18 else return 0;
19 }
20
21 int main()
22 {
23 // freopen("1.in","r",stdin);
24 // freopen("t1.out","w",stdout);
25 int T, cas=1;
26 int n;
27 LL m;
28 while(scanf("%d%lld", &n, &m) == 2)
29 {
30 for(int i=0;i<n;++i)
31 scanf("%lld", &a[i]);
32 sort(a, a+n);
33 int flag = 0;
34 long long sum = m;
35 for(int i=0;i<n&&!flag;++i)
36 {
37 memset(vis, 0, sizeof vis);
38 for(int j=0;j<i;++j) vis[j] = 1;
39 vis[i] = 2;
40 if(sum == a[i])
41 flag = 1;
42 else
43 {
44 flag = check(i, sum, a[i]);
45 if(!flag)
46 {
47 memset(vis, 0, sizeof vis);
48 for(int j=0;j<i;++j) vis[j] = 1;
49 flag = check(i, sum, 0);
50 }
51 sum += a[i];
52 }
53 }
54 printf("Case %d: ",cas++);
55 puts(flag?"YES":"NO");
56 /* if(flag)
57 {
58 LL sum = m;
59 for(int i=0;i<n;++i) if(vis[i] == 1) printf("%lld ", a[i]), sum += a[i]; puts("");
60 printf("lsum:%lld\n", sum);
61 sum = 0;
62 for(int i=0;i<n;++i) if(vis[i] == 2) printf("%lld ", a[i]), sum += a[i]; puts("");
63 printf("rsum:%lld\n", sum);
64 }*/
65 }
66
67 // printf("time=%.3lf\n",(double)clock()/CLOCKS_PER_SEC);
68 return 0;
69 }
solve.cpp
题解：
由砝码质量的上限可知n只有几十。
若a[i]刚好大于m，则不会用到a[i+1],
因为a[1]+a[2]+...+a[i]+m < a[i+1]/2+m <= a[i+1]/2+a[i+1]/3 < a[i+1]
递归执行以下过程
1.每次搜索比m小的a[i]能否凑成m
2.若不可以凑成，则m=a[i]-m,回到1
3.若2还是不可以凑成，则m=m-a[i-1],回到1
搜索能否凑成a[i]时记得剪枝
时间： 2024-08-25 16:02:20