OpenJ_Bailian - 3151 Pots BFS找最短路

1.题意：给你两个容量为A和B的杯子，让你进行如下操作：

1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i)      empty the pot i to the drain;
3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot jis full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j)

求其中一个杯子达到C容量水时最小的操作步数。

2.分析：

求最小步数，每次可以进行六种操作，转化为BFS；

这里吧两个杯子此时刻的状态作为一个元素存入队列！比较抽象运用BFS到具体实际问题里，这个是很难想的也是这个题目的亮点

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;
int A,B,C;
struct Node
{
    int id,v1,v2;//此状态下的步数和第一个容器的水量和第二个容器的水量
    vector<int> op;//保存操作
    //Node(int i,int a,int b):id(i),v1(a),v2(b) {op.clear();}
};
bool judge[105][105];
char oper[6][10] = {"FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
bool BFS()
{
    queue<Node> que;
    Node q;
    q.id = 0;
    q.v1 = 0;
    q.v2 = 0;
    q.op.clear();
    que.push(q);
    while(!que.empty()){
        Node node = que.front();
        que.pop();
        if(node.v1==C||node.v2==C){//输出
            printf("%d\n",node.id);
            for(int i = 0;i<node.op.size();i++){
                printf("%s\n",oper[node.op[i]]);
            }
            return true;
        }
        for(int i = 0;i<6;i++){
            Node another;
            another.id = node.id + 1;
            another.op = node.op;
            another.op.push_back(i);
            if(i==0){//进行六种操作
                if(node.v1==A)continue;
                another.v1 = A;
                another.v2 = node.v2;
                if(!judge[another.v1][another.v2]){
                    judge[another.v1][another.v2] = 1;
                    que.push(another);
                }
            }
            else if(i==1){
                if(node.v2==B)continue;
                another.v1 = node.v1;
                another.v2 = B;
                if(!judge[another.v1][another.v2]){
                    judge[another.v1][another.v2] = 1;
                    que.push(another);
                }
            }
            else if(i==2){
                if(node.v1==0)continue;
                another.v1 = 0;
                another.v2 = node.v2;
                if(!judge[another.v1][another.v2]){
                    judge[another.v1][another.v2] = 1;
                    que.push(another);
                }
            }
            else if(i==3){
                if(node.v2==0)continue;
                another.v1 = node.v1;
                another.v2 = 0;
                if(!judge[another.v1][another.v2]){
                    judge[another.v1][another.v2] = 1;
                    que.push(another);
                }
            }
            else if(i==4){
                if(node.v1==0||node.v2==B)continue;
                int v = B - node.v2;
                if(node.v1>=v){
                    another.v1 = node.v1 - v;
                    another.v2 = B;
                }
                else if(node.v1<v){
                    another.v1 = 0;
                    another.v2 = node.v2 + node.v1;
                }
                if(!judge[another.v1][another.v2]){
                    judge[another.v1][another.v2] = 1;
                    que.push(another);
                }
            }
            else if(i==5){
                if(node.v1==A||node.v2==0)continue;
                int v = A - node.v1;
                if(node.v2>=v){
                    another.v1 = A;
                    another.v2 = node.v2 - v;
                }
                else if(node.v2<v){
                    another.v1 = node.v2 + node.v1;
                    another.v2 = 0;
                }
                if(!judge[another.v1][another.v2]){
                    judge[another.v1][another.v2] = 1;
                    que.push(another);
                }
            }
        }

    }
    return false;
}
int main()
{
    scanf("%d%d%d",&A,&B,&C);
    memset(judge,0,sizeof(judge));
    if(!BFS())printf("impossible\n");
    return 0;
}