Jiudu OJ; question 1147: Jugs, jiudu oj1147jugs

Jiudu OJ; question 1147: Jugs, jiudu oj1147jugs

Original question link address: http://ac.jobdu.com/problem.php? Pid = 1, 1147

Reprinted please indicate this article link: http://blog.csdn.net/yangnanhai93/article/details/42016353

BFS is a simple idea, but pay attention to pruning, because many of them will repeat, for example, the constant empty, this repetition is very serious, so it is necessary to remove duplication, that is, record the matrix of 1000*1000, make sure that the desired a and B are not repeated.

#include <stdio.h>#include <queue>#include <string>#include <memory.h>using namespace std; string op[6]={"fill A","fill B","pour B A","pour A B","empty A","empty B"};bool visited[1001][1001];struct Node{    int left,right;    vector<int> op;};void Cal(int a,int b,int q){    queue<Node> result;    Node first,second;    first.left=0;    first.right=0;    result.push(first);    memset(visited,0,sizeof(visited));    while(!result.empty())    {        first=result.front();        result.pop();        for(int i=0;i<6;i++)        {            second=first;            switch (i)            {            case 0:                second.left=a;                break;            case 1:                second.right=b;                break;            case 2://pour b to a                if(second.right<=a-second.left)                {                    second.left=second.left+second.right;                    second.right=0;                }                else                {                    second.right=second.right-(a-second.left);                    second.left=a;                                  }                break;            case 3:                if(second.left<=b-second.right)                {                    second.right=second.right+second.left;                    second.left=0;                }                else                {                    second.left=second.left-(b-second.right);                    second.right=b;                }                break;            case 4:                second.left=0;                break;            case 5:                second.right=0;                break;            }            second.op.push_back(i);            if(second.right==q)            {                for(int i=0;i<second.op.size();i++)                    printf("%s\n",op[second.op[i]].c_str());                printf("success\n");                return;            }            else            {                if(!visited[second.left][second.right])                    result.push(second);                visited[second.left][second.right]=true;            }        }    }}int main(){    int a,b,q;    while(scanf("%d%d%d",&a,&b,&q)!=EOF)    {        Cal(a,b,q);    }    return 0;}/**************************************************************    Problem: 1147    User: vincent_ynh    Language: C++    Result: Accepted    Time:10 ms    Memory:2036 kb****************************************************************/