poj 3414 pots

来源：互联网发布：linux 编译环境编辑：程序博客网时间：2024/06/05 10:14

题意：有两个容器，一个容量为A,一个容量为B; 有三种操作：
1. FILL（i） : 给容器i装满水，i = 1 或2；
2. DROP（i )：倒掉容器i的所有水；
3. POUR（i，j）：将i中水倒入j中，直到j装满，i中可以有剩余，或者i不足以装满j；

思路： BFS 加上记忆路径，难点在打印路径。

代码很丑。。。

#include "iostream"#include "queue"#include "cstring"using namespace std;int A,B,C;int visit[105][105];  //判断是否访问过此状态struct Node{    int prex;        //记录前一个状态的A容器状态    int prey;       //记录前一个状态的B容器状态    int x;         //x，y为此时状态    int y;        int step;     //步数    int op;       //前一状态到此状态的操作，有六种};Node pot[105][105]; 用于记忆路径void output(Node s) {    //打印路径    if(s.prex == -1 && s.prey == -1) return;    else {        output(pot[s.prex][s.prey]);  //递归，总是再打印前一操作后再打印后一步操作        switch(s.op) {             case  1:cout <<"FILL(1)\n";break;             case  2:cout <<"FILL(2)\n";break;             case  3:cout <<"DROP(1)\n";break;             case  4:cout <<"DROP(2)\n";break;             case  5:cout <<"POUR(1,2)\n";break;             case  6:cout <<"POUR(2,1)\n";break;             default:cout <<"something wrong\n";break;        }    }}void bfs() { //简单的宽搜    Node s;    s.prex = -1;    s.prey = -1;    s.x = 0;    s.y = 0;    s.step = 0;    s.op = -1;    memset(visit,0,sizeof(visit));    queue<Node>q;    while(!q.empty()) q.pop();    q.push(s);    while(!q.empty()) {        s = q.front();        q.pop();        pot[s.x][s.y] = s;  //将此点记录下来        if(s.x == C || s.y == C) {            cout << s.step << endl;            output(s);            return;        }        visit[s.x][s.y] = 1;        for(int i = 1; i <= 6; i++) { //六个操作            if(i == 1 && !visit[A][s.y]) {                visit[A][s.y] = 1;                Node next;                next.prex = s.x;                next.prey = s.y;                next.x = A;                next.y = s.y;                next.step = s.step+1;                next.op = i;                q.push(next);            }            if(i == 2 && !visit[s.x][B]) {                visit[s.x][B] = 1;                Node next;                next.prex = s.x;                next.prey = s.y;                next.x = s.x;                next.y = B;                next.step = s.step+1;                next.op = i;                q.push(next);            }            if(i == 3 && !visit[0][s.y]) {                visit[0][s.y] = 1;                Node next;                next.prex = s.x;                next.prey = s.y;                next.x = 0;                next.y = s.y;                next.step = s.step+1;                next.op = i;                q.push(next);            }            if(i == 4 && !visit[s.x][0]) {                visit[s.x][0] = 1;                Node next;                next.prex = s.x;                next.prey = s.y;                next.x = s.x;                next.y = 0;                next.step = s.step+1;                next.op = i;                q.push(next);            }            if(i == 5 ) {                int nx,ny;                if(s.x + s.y > B) {                     nx = s.x + s.y - B;                     ny = B;                }                else if(s.x+s.y <= B) {                    nx = 0;                    ny = s.x + s.y;                }                if(!visit[nx][ny]) {                    visit[nx][ny] = 1;                    Node next;                    next.prex = s.x;                    next.prey = s.y;                    next.x = nx;                    next.y = ny;                    next.step = s.step+1;                    next.op = i;                    q.push(next);                }            }            if(i == 6) {                int nx,ny;                if(s.x + s.y > A) {                     ny = s.x + s.y - A;                     nx = A;                }                else if(s.x+s.y <= A) {                    ny = 0;                    nx = s.x + s.y;                }                 if(!visit[nx][ny]) {                    visit[nx][ny] = 1;                     Node next;                    next.prex = s.x;                    next.prey = s.y;                    next.x = nx;                    next.y = ny;                    next.step = s.step+1;                    next.op = i;                    q.push(next);                }            }        }    }    cout << "impossible" << endl;    return;}int main() {    while(cin >> A >> B >> C) {        bfs();    }return 0;}

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