bfs + 路径输出

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
        DROP(i)      empty the pot i to the drain;
        POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is 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).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input

3 5 4

Sample Output

6
    FILL(2)
    POUR(2,1)
    DROP(1)
    POUR(2,1)
    FILL(2)
    POUR(2,1)

题意 ：

3 种操作，第一种将桶内装满，第二种将桶倒干净，第三种将一个桶内的倒入另一个桶。

思路 ：

BFS做，每次将当前的状态打入队列，最后要输出路径，在结构体中存上一个点的位置。

代码 ：

int a, b, c;
struct node
{
    int x, y;
    int pt;
    int temp;
    node *pre;
}arr[400];

int vis[105][105];
int ans;
stack<int>s;
void bfs(){
    queue<node>que; // 每次将当前的状态打入队列，对于每个状态有 6 种处理方法
    memset(vis, 0, sizeof(vis));

    node v;
    v.x = 0, v.y = 0;
    v.pt = 0;
    v.temp = 0;
    v.pre = NULL;
    while(!que.empty()) que.pop();
    que.push(v);
    int cnt = -1;
    vis[0][0] = 1;

    while(!que.empty()){
        v = que.front();
        que.pop();
        cnt++;
        arr[cnt] = v;

        for(int i = 1; i <= 6; i++){
            switch(i){
                case 1:
                    v.x = a;
                    v.y = arr[cnt].y;
                    v.pt = 1;
                    break;
                case 2:
                    v.x = arr[cnt].x;
                    v.y = b;
                    v.pt = 2;
                    break;
                case 3:
                    v.x = 0;
                    v.y = arr[cnt].y;
                    v.pt = 3;
                    break;
                case 4:
                    v.x = arr[cnt].x;
                    v.y = 0;
                    v.pt = 4;
                    break;
                case 5:
                    if (arr[cnt].x > (b - arr[cnt].y)){
                        v.x = arr[cnt].x - (b - arr[cnt].y);
                        v.y = b;
                    }
                    else {
                        v.x = 0;
                        v.y = arr[cnt].x + arr[cnt].y;
                    }
                    v.pt = 5;
                    break;
                case 6:
                    if (arr[cnt].y > (a - arr[cnt].x)){
                        v.x = a;
                        v.y = arr[cnt].y - (a - arr[cnt].x);
                    }
                    else {
                        v.x = arr[cnt].x + arr[cnt].y;
                        v.y = 0;
                    }
                    v.pt = 6;
            }
            if (vis[v.x][v.y]) continue;
            vis[v.x][v.y] = 1;
            v.temp = arr[cnt].temp + 1;
            v.pre = &arr[cnt];
            if (v.x == c || v.y == c){

                while(!s.empty()) s.pop();
                ans = v.temp;
                while(v.pre){
                    s.push(v.pt);
                    v = *v.pre;
                }
                return;
            }
            que.push(v);
        }
    }
}

void print(){
    while(!s.empty()){
        int f = s.top();
        s.pop();
        switch(f){
            case 1: printf("FILL(1)\n");break;
            case 2: printf("FILL(2)\n");break;
            case 3: printf("DROP(1)\n");break;
            case 4: printf("DROP(2)\n");break;
            case 5: printf("POUR(1,2)\n");break;
            case 6: printf("POUR(2,1)\n");break;
        }
    }
}

int main() {

    while(~scanf("%d%d%d", &a, &b, &c)){
        ans = 0;
        bfs();

        if (ans == 0) printf("impossible\n");
        else {
            printf("%d\n", ans);
            print();
        }
    }
    return 0;
}