（beginer）最短路 UVA 10603 Fill

Problem D

Fill

 

There are three jugs with a volume of a, b and c liters. (a, b, and c are positive integers not greater than 200). The first and the second jug are initially empty, while the third

is completely filled with water. It is allowed to pour water from one jug into another until either the first one is empty or the second one is full. This operation can be performed zero, one or more times.

 

You are to write a program that computes the least total amount of water that needs to be poured; so that at least one of the jugs contains exactly d liters of water (d is a positive integer not greater than 200). If it is not possible to measure d liters this way your program should find a smaller amount of water d' < d which is closest to d and for which d' liters could be produced. When d' is found, your program should compute the least total amount of poured water needed to produce d' liters in at least one of the jugs.

 

Input

The first line of input contains the number of test cases. In the next T lines, T test cases follow. Each test case is given in one line of input containing four space separated integers - a, b, c and d.

 

Output

The output consists of two integers separated by a single space. The first integer equals the least total amount (the sum of all waters you pour from one jug to another) of poured water. The second integer equals d, if d liters of water could be produced by such transformations, or equals the closest smaller value d' that your program has found.

 

Sample Input	Sample Output
2 2 3 4 2 96 97 199 62	2 2 9859 62

题意：有三个杯子，容量一定，一开始第一二个杯子是空的，第三个杯子是满的，然后就把水倒到别的杯子，问最后能否达到最少有一个杯子里面的水有d升，问需要倒的水的数量最少多少。如果不能达到，小于d的又是最大的是多少，最少到多少数量的水能到达这个状态。

思路：我们用一个dist[a][b][c]记录到达这个状态的最小“距离”，然后就dfs就行了。

代码：

#include<iostream>

#include<queue>

#include<stdio.h>

#include<cstring>

#include<string.h>

#include<queue>

using namespace std;

const int maxn = 205;

struct State

{

State(int aa,int bb,int cc,int dd)

:a(aa) , b(bb) , c(cc) , d(dd) { }

int a , b, c;

int d;

};

bool operator < (const State & s1 , const State & s2)

{

return s1.d < s2.d;

}

int dist[maxn][maxn][maxn];

priority_queue<State> q;

int a , b , c , d;

int ans_d , ans_cost;

void dfs(int aa,int bb,int cc,int step)

{

if (dist[aa][bb][cc] <= step) return;

if (aa==d || bb==d || cc==d)

{

if (ans_d!=d) {

ans_d = d;

ans_cost = step;

}

else if (ans_cost > step) 

ans_cost = step;

}

if (aa < d && aa >= ans_d) { 

if (aa==ans_d) { 

if (ans_cost > step) 

ans_cost = step;

} else {

ans_cost = step;

ans_d = aa;

}

}

if (bb < d && bb >= ans_d) { 

if (bb==ans_d) {

if (ans_cost > step)

ans_cost = step;

} else {

ans_cost = step;

ans_d = bb;

}

}

if (cc < d && cc >= ans_d) { 

if (cc==ans_d) {

if (ans_cost > step)

ans_cost = step;

} else {

ans_cost = step;

ans_d = cc;

}

}

dist[aa][bb][cc] = step;

int cost;

if (aa > 0) {

cost = min(b-bb,aa);

if (bb < b) dfs(aa-cost,bb+cost,cc,step+cost);

cost = min(c-cc,aa);

if (cc < c) dfs(aa-cost,bb,cc+cost,step+cost);

}

if (bb > 0) {

cost = min(a-aa,bb);

if (aa < a) dfs(aa+cost,bb-cost,cc,step+cost);

cost = min(c-cc,bb);

if (cc < c) dfs(aa,bb-cost,cc+cost,step+cost);

}

if (cc > 0) {

cost = min(a-aa,cc);

if (aa < a) dfs(aa+cost,bb,cc-cost,step+cost);

cost = min(b-bb,cc);

if (bb < b) dfs(aa,bb+cost,cc-cost,step+cost);

}

}

int main()

{

int T; cin>>T;

while (T--)

{

scanf("%d%d%d%d",&a,&b,&c,&d);

memset(dist,0x3f,sizeof(dist));

ans_cost = 1e9 , ans_d = 0;

dfs(0,0,c,0);

printf("%d %d\n",ans_cost,ans_d);

}

}