POJ-3414.Pots.(BFS + 路径打印)

　　这道题做了很长时间，一开始上课的时候手写代码，所以想到了很多细节，但是创客手打代码的时候由于疏忽又未将pair赋初值，导致一直输出错误，以后自己写代码可以专心一点，可能会在宿舍图书馆或者Myhome，创客晚上好吵呀，隔壁真的服...

　　本题大意：给定两个杯子的容量，有六种操作，通过操作使得两个被子其中一个杯子的水等于待输入值C，并输出最少操作次数和操作名称。

　　本题思路：BFS，遇到合适的直接输出，搜完都没有搜到就impossible。

　　参考代码：

 　　将switch改为if能节省大部分篇幅，但是switch更美观明了一点。

#include <string>
#include <cstring>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;

typedef pair<int ,int > P;
const int maxn = 100 + 5, INF = 0x3f3f3f3f;
int A, B, C, t;

struct {
    int x, y, cnt;//用来存他爹的坐标和倒水的方式
} Ans[maxn][maxn];
int vis[maxn][maxn];
string ans[6] = {
    "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)",
};

void bfs() {
    queue <P> Q;
    vis[0][0] = 0;
    Ans[0][0].x = -1, Ans[0][0].y = -1;
    Q.push(make_pair(0, 0));
    while(!Q.empty()) {
        P now = Q.front(), Next;
        stack <string> s1;
        Q.pop();
        if(now.first == C || now.second == C) {
            cout << vis[now.first][now.second] << endl;
            int X = now.first;
            int Y = now.second;
            while(X != -1) {
                s1.push(ans[Ans[X][Y].cnt]);
                int tmp = X;
                X = Ans[X][Y].x;
                Y = Ans[tmp][Y].y;
            }
            s1.pop();
            while(!s1.empty()) {
                cout << s1.top() << endl;
                s1.pop();
            }
            return;
        }
        for(int i = 0; i < 6; i ++) {
            switch(i) {
                case 0 ://将A加满
                   Next.first = A;
                   Next.second = now.second;
                   break;
                case 1 ://将B加满
                    Next.second = B;
                    Next.first = now.first;
                    break;
                case 2 ://将A倒掉
                    Next.second = now.second;
                    Next.first = 0;
                    break;
                case 3 ://将B倒掉
                    Next.first = now.first;
                    Next.second = 0;
                    break;
                case 4 ://将A倒入B中
                    t = B - now.second;//B中还能倒入的量
                    Next.first = now.first - t;
                    if(Next.first < 0) Next.first = 0;
                    Next.second = now.first + now.second;
                    if(Next.second > B) Next.second = B;
                    break;
                case 5 ://将B倒入A中
                    t = A - now.first;//A中还能倒入的量
                    Next.second = now.second  - t;
                    if(Next.second < 0) Next.second = 0;
                    Next.first = now.first + now.second;
                    if(Next.first > A) Next.first = A;
                    break;
            }
            if(vis[Next.first][Next.second] == INF) {
                vis[Next.first][Next.second] = vis[now.first][now.second] + 1;
                Q.push(Next);
                Ans[Next.first][Next.second].x = now.first;
                Ans[Next.first][Next.second].y = now.second;
                Ans[Next.first][Next.second].cnt = i;
            }
        }
    }
    cout << "impossible" << endl;
}

int main () {
    memset(vis, INF, sizeof(vis));
    cin >> A >> B >> C;
    bfs();
    return 0;
}