uva10603

/*
solution:
还是把状态看成结点，然后以隐式图的方法去扩展结点，其过程跟dijkstra算法有点类似。

note:
隐式图

date:
2016/5/4
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;
const int maxn = 200 + 5;

int vis[maxn][maxn], ans[maxn], cap[3], d;  //ans[i]代表在这个杯子里有i量的水，转移了多少量的水

struct Node {
int v[3], dist;

bool operator < (const Node& rhs) const {
return dist > rhs.dist;
}
};

int main()
{
//freopen("input.txt", "r", stdin);
int kase;
scanf("%d", &kase);
while(kase--) {
scanf("%d%d%d%d", &cap[0], &cap[1], &cap[2], &d);
memset(vis, 0 ,sizeof(vis));
memset(ans, -1, sizeof(ans));

Node start;
start.dist = 0;
start.v[0] = start.v[1] = 0;
start.v[2] = cap[2];    //初始化开始结点

priority_queue<Node> pq;
pq.push(start);
vis[0][0] = 1;

while(!pq.empty()) {
Node x = pq.top();  pq.pop();

for(int i = 0; i < 3; i++) {
int d = x.v[i];
if(ans[d] < 0 || x.dist < ans[d])   ans[d] = x.dist;    //更新ans数组中的值
}

if(ans[d] >= 0) break;

//扩展节点，从3个杯子里选出两个，把其中一个倒进另外一个
for(int i = 0; i < 3; i++)  //把i号杯子里的水倒进j号杯子里
for(int j = 0; j < 3; j++)
if(i != j) {
if(x.v[i] == 0 || x.v[j] == cap[j]) continue;
int amount = min(cap[j], x.v[i] + x.v[j]) - x.v[j]; //求出最小应该倒多少水

//扩展结点
Node x2;
memcpy(&x2, &x, sizeof(x));
x2.v[i] -= amount;
x2.v[j] += amount;
x2.dist = x.dist + amount;

if(!vis[x2.v[0]][x2.v[1]]) {
vis[x2.v[0]][x2.v[1]] = 1;
pq.push(x2);
}
}
}

while(d >= 0) {
if(ans[d] >= 0) {
printf("%d %d\n", ans[d], d);
break;
}
d--;
}
}
return 0;
}