你的任务是写一个程序计算出用最少的倒水量,使得其中一个杯子里有d升水。如果不能倒出d升水的话,那么找到一个d' < d ,使得d' 最接近d。
【解题方法】 本题最多的状态数不超过(A+1)*(B+1),故BFS所有状态即可,代码参考: 紫书代码仓库。具体讲解可以看紫书P202页,讲得很细节和清楚。
【AC代码】
//
//Created by BLUEBUFF 2016/1/6
//Copyright (c) 2016 BLUEBUFF.All Rights Reserved
//
#pragma comment(linker,"/STACK:102400000,102400000")
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
//using namespace __gnu_pbds;
typedef long long LL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b) memset(a, b, sizeof(a))
#define MP(x, y) make_pair(x,y)
const int maxn = 220;
const int maxm = 2e5;
const int maxs = 10;
const int maxp = 1e3 + 10;
const int INF = 1e9;
const int UNF = -1e9;
const int mod = 1e9 + 7;
int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;
//head
struct node{
int v[3], dist;
bool operator < (const node &rhs) const{
return dist > rhs.dist;
}
};
int vis[maxn][maxn], cap[3], ans[maxn];
void update_ans(const node &u){
for(int i = 0; i < 3; i++){
int d = u.v[i];
if(ans[d] < 0 || u.dist < ans[d]) ans[d] = u.dist;
}
}
void solve(int a, int b, int c, int d){
cap[0] = a, cap[1] = b, cap[2] = c;
memset(vis, 0, sizeof(vis));
memset(ans, -1, sizeof(ans));
priority_queue <node> q;
node st;
st.dist = 0;
st.v[0] = 0, st.v[1] = 0, st.v[2] = c;
q.push(st);
vis[0][0] = 1;
while(!q.empty()){
node u = q.top(); q.pop();
update_ans(u);
if(ans[d] >= 0) break;
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
if(i != j){
if(u.v[i] == 0 || u.v[j] == cap[j]) continue;
int amount = min(cap[j], u.v[i] + u.v[j]) - u.v[j];
node u2;
memcpy(&u2, &u, sizeof(u));
u2.dist = u.dist + amount;
u2.v[i] -= amount;
u2.v[j] += amount;
if(!vis[u2.v[0]][u2.v[1]]){
vis[u2.v[0]][u2.v[1]] = 1;
q.push(u2);
}
}
}
}
}
while(d >= 0){
if(ans[d] >= 0){
printf("%d %d\n", ans[d], d);
return ;
}
d--;
}
}
int main()
{
int T, a, b, c, d;
scanf("%d", &T);
while(T--){
scanf("%d%d%d%d", &a, &b, &c, &d);
solve(a, b, c, d);
}
return 0;
}