Milk Pails

Milk Pails"/>

Milk Pails

【题目原文】
Farmer John has received an order for exactly M units of milk (1≤M≤200) that he needs to fill right away. Unfortunately, his fancy milking machine has just become broken, and all he has are two milk pails of integer sizes X and Y (1≤X,Y≤100) with which he can measure milk. Both pails are initially empty. Using these two pails, he can perform up to K of the following types of operations (1≤K≤100):
- He can fill either pail completely to the top.
- He can empty either pail.
- He can pour the contents of one pail into the other, stopping when the former becomes empty or the latter becomes full (whichever of these happens first).
Although FJ realizes he may not be able to end up with exactly M total units of milk in the two pails, please help him compute the minimum amount of error between M and the total amount of milk in the two pails. That is, please compute the minimum value of |M−M′| such that FJ can construct M′ units of milk collectively between the two pails.
【题目描述】
有两个桶大小为X和Y，开始都是空的。
有如下三种操作，最多操作K次。
①倒满一个桶。
②倒空一个桶。
③把一个桶里的油倒到另一个桶里，直到一个桶空了或者另一个桶满了。（看哪种情况先发生）
给出要求的油量M.
最后你获得的油量是两个桶的水量之和M’。
求|M-M’|的最小值。
【输入格式】
一行，四个数X,Y,K,M
【输出格式】
最小值。
【样例输入】
14 50 2 32
【样例输出】
18
【数据范围】
1<=x,y,k<=1000
1<=m<=2000
【分析】
此题有多种方法。
方法1：Bool型DP，设f[i][j][k]表示第i步时一个桶为i，一个桶为j能否行得通（0\1），最后打擂台即可。程序较简单，容易实现，但是时间和空间都不资瓷，只能做到百级别的数据。

#include<bits/stdc++.h>
using namespace std;
int f[105][105][105];
int abs(int x){ return (x<0?-x:x); }
int main(){int x,y,k,m;cin>>x>>y>>k>>m;memset(f,0,sizeof(f));f[1][x][0]=f[1][0][y]=1;int ans=(1<<31)-1;for (int n=1;n<=k;n++)for (int i=0;i<=x;i++)for (int j=0;j<=y;j++){if (f[n][i][j]) {ans=min(ans,abs(m-i-j));f[n+1][0][j]=f[n+1][i][0]=f[n+1][x][j]=f[n+1][i][y]=1;if (i+j<=y) f[n+1][0][i+j]=1; else f[n+1][i-(y-j)][y]=1;if (i+j<=x) f[n+1][i+j][0]=1; else f[n+1][x][j-(x-i)]=1;}}cout<<ans;
}

方法2：搜索（深搜广搜都可以），设f[i][j]表示一个桶为i，一个桶为j的最少步数，这下可以做到千级别的数据了。

#include<bits/stdc++.h>
using namespace std;
int x,y,k,m,ans;
int a[2005][2005]; 
queue<int> a1,a2;
int main(){cin>>x>>y>>k>>m;k++;a1.push(0);a2.push(0);a[0][0]=1;ans=m;while(!a1.empty()){int t1=a1.front(),t2=a2.front();a1.pop();a2.pop();if (abs(t1+t2-m)<ans) ans=abs(t1+t2-m);if (a[t1][t2]==k) continue;if (!a[x][t2]){a[x][t2]=a[t1][t2]+1;a1.push(x);a2.push(t2);}if (!a[t1][y]){a[t1][y]=a[t1][t2]+1;a1.push(t1);a2.push(y);}if (!a[0][t2]){a[0][t2]=a[t1][t2]+1;a1.push(0);a2.push(t2);}if (!a[t1][0]){a[t1][0]=a[t1][t2]+1;a1.push(t1);a2.push(0);}if (t1+t2<=y&&!a[0][t1+t2]){a[0][t1+t2]=a[t1][t2]+1;a1.push(0);a2.push(t1+t2);}if (t1+t2<=x&&!a[t1+t2][0]){a[t1+t2][0]=a[t1][t2]+1;a1.push(t1+t2);a2.push(0);}if (t1>y-t2&&!a[t1+t2-y][y]){a[t1+t2-y][y]=a[t1][t2]+1;a1.push(t1+t2-y);a2.push(y);}if (t2>x-t1&&!a[x][t1+t2-x]){a[x][t1+t2-x]=a[t1][t2]+1;a1.push(x);a2.push(t1+t2-x);}if (ans==0){cout<<0;return 0;}}cout<<ans;
}