POJ3414 Pots —— BFS + 模拟

题目链接：http://poj.org/problem?id=3414

Pots

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18550 Accepted: 7843 Special Judge

Description

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

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 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

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

Source

Northeastern Europe 2002, Western Subregion

题解：

类似的题目，也是有关几个容器倒来倒去，然后求一个目标量：http://blog.csdn.net/dolfamingo/article/details/77804030

此题不过是多了个路径输出，队列把STL换成数组就可以了，不细述。

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <string>#include <set>#define ms(a,b) memset((a),(b),sizeof((a)))using namespace std;typedef long long LL;const int INF = 2e9;const LL LNF = 9e18;const int MOD = 1e9+7;const int MAXN = 100+10;struct node{    //con[i]为容器i当前的量；op存操作的信息；pre为上一步所在队列的下标（为了输出路径）    int con[2], op[3], step, pre;};int vis[MAXN][MAXN], vol[3];    //vol[0]、vol[1]为两个容器的容量， vol[3]为目标量node que[100010];   //由于要输出路径，就要用数组来作队列了int front, rear;int bfs(){    ms(vis,0);    front = rear = 0;    node now, tmp;    now.con[0] = now.con[1] = 0;    now.step = 0;    vis[0][0] = 1;    que[rear++] = now;    while(front!=rear)    {        now = que[front++];        if(now.con[0]==vol[2] || now.con[1]==vol[2])            return front-1;        for(int i = 0; i<2; i++)        {            tmp = now;      //操作1：FILL            tmp.con[i] = vol[i];            if(!vis[tmp.con[0]][tmp.con[1]])            {                vis[tmp.con[0]][tmp.con[1]] = 1;                tmp.op[0] = 1; tmp.op[1] = i;                tmp.step = now.step+1;                tmp.pre = front-1;                que[rear++] = tmp;            }            tmp = now;      //操作2：DROP            tmp.con[i] = 0;            if(!vis[tmp.con[0]][tmp.con[1]])            {                vis[tmp.con[0]][tmp.con[1]] = 1;                tmp.op[0] = 2; tmp.op[1] = i;                tmp.step = now.step+1;                tmp.pre = front-1;                que[rear++] = tmp;            }            tmp = now;      //操作三POUR            int j = !i; //另外一个容器            int pour = min(tmp.con[i], vol[j]-tmp.con[j]);            tmp.con[j] += pour;            tmp.con[i] -= pour;            if(!vis[tmp.con[0]][tmp.con[1]])            {                vis[tmp.con[0]][tmp.con[1]] = 1;                tmp.op[0] = 3; tmp.op[1] = i; tmp.op[2] = j;                tmp.step = now.step+1;                tmp.pre = front-1;                que[rear++] = tmp;            }        }    }    return -1;}void Print(int i)   //输出路径{    if(que[i].pre!=0)        Print(que[i].pre);    if(que[i].op[0]==1)        printf("FILL(%d)\n", que[i].op[1]+1);    else if(que[i].op[0]==2)        printf("DROP(%d)\n", que[i].op[1]+1);    else        printf("POUR(%d,%d)\n", que[i].op[1]+1, que[i].op[2]+1);}int main(){    while(scanf("%d%d%d",&vol[0], &vol[1], &vol[2])!=EOF)    {        int i = bfs();        if(i==-1)            puts("impossible");        else        {            printf("%d\n",que[i].step);            Print(i);        }    }}