J - 砝码称重 改自51nod1449

Posted 2020-09-22 dy blog

J - 砝码称重

Time Limit: 2000/1000 MS (Java/Others)      Memory Limit: 128000/64000 KB (Java/Others)

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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