P2214 [USACO14MAR]哞哞哞Mooo Moo

Farmer John has completely forgotten how many cows he owns! He is too embarrassed to go to his fields to count the cows, since he doesn’t want the cows to realize his mental lapse. Instead, he decides to count his cows secretly by planting microphones in the fields in which his cows tend to gather, figuring that he can determine the number of cows from the total volume of all the mooing he hears.

FJ’s N fields (1 <= N <= 100) are all arranged in a line along a long straight road. Each field might contain several types of cows; FJ owns cows that come from B different breeds (1 <= B <= 20), and a cow of breed i moos at a volume of V(i) (1 <= V(i) <= 100). Moreover, there is a strong wind blowing down the road, which carries the sound of mooing in one direction from left to right: if the volume of mooing in some field is X, then in the next field this will contribute X-1 to the total mooing volume (and X-2 in the field after that, etc.). Otherwise stated, the mooing volume in a field is the sum of the contribution due to cows in that field, plus X-1, where X is the total mooing volume in the preceding field.

Given the volume of mooing that FJ records in each field, please compute the minimum possible number of cows FJ might own.

The volume FJ records in any field is at most 100,000.

民约翰忘记了他到底有多少头牛，他希望通过收集牛叫声的音量来计算牛的数量。

他的N (1 <= N <= 100)个农场分布在一条直线上，每个农场可能包含B (1 <= B <= 20)个品种的牛，一头品种i的牛的音量是V(i) ，(1 <= V(i) <= 100)。一阵大风将牛的叫声从左往右传递，如果某个农场的总音量是X，那么将传递X-1的音量到右边的下一个农场。另外，一个农场的总音量等于该农场的牛产生的音量加上从上一个农场传递过来的音量（即X-1）。任意一个农场的总音量不超过100000。

请计算出最少可能的牛的数量。

输入格式：

• Line 1: The integers N and B.

• Lines 2..1+B: Line i+1 contains the integer V(i).

• Lines 2+B..1+B+N: Line 1+B+i contains the total volume of all mooing

in field i.

输出格式：

• Line 1: The minimum number of cows owned by FJ, or -1 if there is no

configuration of cows consistent with the input.

输入样例#1： 复制

5 2
5
7
0
17
16
20
19

输出样例#1： 复制

4

INPUT DETAILS:

FJ owns 5 fields, with mooing volumes 0,17,16,20,19. There are two breeds

of cows; the first moos at a volume of 5, and the other at a volume of 7.

OUTPUT DETAILS:

There are 2 cows of breed #1 and 1 cow of breed #2 in field 2, and there is

another cow of breed #1 in field 4.

Source: USACO 2014 March Contest, Silver

• 注意到声音只会从左往右传,那么就可以推出每一块真正发出的声音b[i]

$$b[i]=a[i]-max(a[i-1]-1,0)$$

• 接下来就是,简单的完全背包

  • 把牛的叫声看作体积,把音量看作总容量,把牛的数量看作商品数

  #include
  #include
  #include
  #include
  #include
  #include
  #define re register int
  using namespace std;
  inline int read(){
  int x=0,w=1;
  char ch=getchar();
  while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
  if(ch=='-') w=-1,ch=getchar();
  while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-48,ch=getchar();
  return x*w;
  }
  const int INF=2147483647;
  int n,k,maxb=-1,ans=0;
  int v[25],a[105],b[105],f[100005];
  int main() {
  freopen("p2214.in","r",stdin);
  freopen("p2214.out","w",stdout);
  n=read(),k=read();
  for(re i=1;i<=k;++i) v[i]=read();
  for(re i=1;i<=n;++i) a[i]=read();
  for(re i=1;i<=n;++i) { b[i]=a[i]-max(a[i-1]-1,0); maxb=max(maxb,b[i]); }
  for(re i=1;i<=maxb;++i) f[i]=INF;
  for(re i=1;i<=k;++i) for(re j=v[i];j<=maxb;j++) { if(f[j-v[i]]!=INF) f[j]=min(f[j],f[j-v[i]]+1); }
  for(re i=1;i<=n;i++) { if(f[b[i]]==INF) { printf("-1\n"); return 0;} else ans+=f[b[i]]; }
  printf("%d\n",ans);
  return 0;
  }

​