Milk Pails
Milk Pails
题目描述
- 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.
输入
输出
样例输入
14 50 2 32
样例输出
18
分析:枚举所有可能情况,bfs即可;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e3+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,x,y,k,mi,vis[maxn][maxn];
void bfs()
{
queue<pii >p;p.push({,});vis[][]=;
while(!p.empty())
{
int u=p.front().fi,v=p.front().se;
p.pop();
mi=min(mi,abs(u+v-m));
if(vis[u][v]==k+)continue;
if(u<x&&!vis[x][v])p.push({x,v}),vis[x][v]=vis[u][v]+;
if(v<y&&!vis[u][y])p.push({u,y}),vis[u][y]=vis[u][v]+;
if(u<x&&v)
{
int mi=min(x-u,v);
if(!vis[u+mi][v-mi])p.push({u+mi,v-mi}),vis[u+mi][v-mi]=vis[u][v]+;
}
if(v<y&&u)
{
int mi=min(y-v,u);
if(!vis[u-mi][v+mi])p.push({u-mi,v+mi}),vis[u-mi][v+mi]=vis[u][v]+;
}
if(u&&!vis[][v])p.push({,v}),vis[][v]=vis[u][v]+;
if(v&&!vis[u][])p.push({u,}),vis[u][]=vis[u][v]+;
}
}
int main()
{
int i,j,t;
scanf("%d%d%d%d",&x,&y,&k,&m);
mi=m;
bfs();
printf("%d\n",mi);
//system ("pause");
return ;
}
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