J - 砝码称重 改自51nod1449
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 }
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 }
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
