SPOJ Pouring Water

最新推荐文章于 2025-06-15 10:05:58 发布

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原文链接：http://www.cnblogs.com/Patt/p/4824875.html

本文通过BFS算法解决了给定两个水罐容量的情况下，如何通过一系列操作获得指定水量的问题。详细阐述了BFS算法的实现过程，包括状态转移、避免重复入队等关键点，并提供了样例代码。

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POUR1 - Pouring water

#gcd #recursion

Given two vessels, one of which can accommodate a litres of water and the other - b litres of water, determine the number of steps required to obtain exactly c litres of water in one of the vessels.

At the beginning both vessels are empty. The following operations are counted as 'steps':

emptying a vessel,
filling a vessel,
pouring water from one vessel to the other, without spilling, until one of the vessels is either full or empty.

Input

An integer t, 1<=t<=100, denoting the number of testcases, followed by t sets of input data, each consisting of three positive integers a, b, c, not larger than 40000, given in separate lines.

Output

For each set of input data, output the minimum number of steps required to obtain c litres, or -1 if this is impossible.

Example

Sample input:

Sample output:

2
-1
------------------------

Solution
BFS
写BFS最重要的是避免同一状态重复入队。
另外检查目标状态应该在每个状态出队时进行，因为状态的出口是“唯一”的，而入口一般有多种情况（即由队首状态一般可转移到多个新状态），注意代码中加粗的那行。
另外由于问题中由初始状态可转移到的状态并不多（也由于二维数组开不下），应当用map存每个状态到初始状态的距离（及所需的最少转移步数）。
还有一个技巧就是将enqueue写成一个函数，这样就避免了向多个状态转移带来的代码重复。

#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int> P;
int gcd(int a, int b){return b?gcd(b, a%b):a;}
map<P,int> mp;
queue<P> que;
void enque(int a, int b, int d){
    int tmp=mp[{a, b}];
    if(!tmp||tmp>d){
        mp[{a, b}]=d;
        que.push({a, b});
    }
}
bool ok(int a, int b, int c){return a==c||b==c;}
// how BFS works?
void bfs(int x, int y, int a, int b, int c){
    mp.clear();
    while(!que.empty()) que.pop();
    mp[{x, y}]=1;
    que.push({x, y});
    while(!que.empty()){
        P top=que.front(); que.pop(); int d=mp[top];
        int x=top.first, y=top.second;
        if(x==c||y==c){printf("%d\n", d-1); return;}
        if(x) enque(0, y, d+1);
        if(y) enque(x, 0, d+1);
        if(x!=a){
            enque(a, y, d+1);
            if(y){
                int add=min(y, a-x);
                 enque(x+add, y-add, d+1);
            }
        }
        if(y!=b){
            enque(x, b, d+1);
            if(x){
                int add=min(x, b-y);
                enque(x-add, y+add, d+1);
            }
        }
    }  
    puts("-1");
}
int main(){
    int T; scanf("%d", &T);
    for(int a, b, c; T--;){
        scanf("%d%d%d", &a, &b, &c);
        if(c>max(a, b)){puts("-1"); continue;}
        if(c==a||c==b){puts("1"); continue;}
        if(a==b){puts("-1"); continue;}
        if(c%gcd(a,b)){puts("-1"); continue;}
        bfs(0, 0, a, b, c);
    }
}