POJ 3414 Pots

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题目大意：
有两个容量分别为a,b的杯子，需要经过若干次变换使任意一个杯子的容量为c，求最短的变换路径
#include <iostream>#include <queue>#include <cstring>using namespace std;int a,b,k;int vis[105][105];struct node{    int x,y,step;    string path;};string p[] = {"FILL(2)","FILL(1)","POUR(2,1)","POUR(1,2)","DROP(1)","DROP(2)"};void print(node cur){    cout << cur.step << endl;    for (int i = 0; i < cur.path.size(); ++i)        cout << p[cur.path[i]-'0'] << endl;}int BFS(){    node cur,next;    cur.x = 0;    cur.y = 0;    cur.step = 0;    cur.path = "";    queue <node> q;    q.push(cur);    while(!q.empty())    {        cur = q.front();        q.pop();        if (cur.x == k|| cur.y == k)        {            print(cur);            return 0;        }       /*fill*/        if (cur.x != a)        {            next = {a,cur.y,cur.step+1};            if (!vis[a][cur.y])            {                next.path = cur.path + "1";                q.push(next);                vis[a][cur.y] = 1;            }        }        if (cur.y != b)        {            next = {cur.x,b,cur.step+1};            if (!vis[cur.x][b])            {                next.path = cur.path + "0";                q.push(next);                vis[cur.x][b] = 1;            }        }        /*drop*/        next = {0,cur.y,cur.step+1};        if(!vis[0][cur.y])        {            next.path = cur.path + "4";            q.push(next);            vis[0][cur.y] = 1;        }        next = {cur.x,0,cur.step+1};        if (!vis[cur.x][0])        {            next.path = cur.path + "5";            q.push(next);            vis[cur.x][0] = 1;        }        /*pour*/        if (cur.x != a && cur.y != 0)        {            int t1 = cur.x, t2 = cur.y;            if (a - t1 <= t2)            next = {a,t2-a+t1,cur.step+1};            else            next = {t1+t2,0,cur.step+1};            if (!vis[next.x][next.y])            {                next.path = cur.path + "2";                vis[next.x][next.y] = 1;                q.push(next);            }        }        if (cur.y != b && cur.x != 0)        {            int t1 = cur.x, t2 = cur.y;            if (b - t2 <= t1)            next = {t1-b+t2,b,cur.step+1};            else            next = {0,t1+t2,cur.step+1};            if (!vis[next.x][next.y])            {                next.path = cur.path + "3";                vis[next.x][next.y] = 1;                q.push(next);            }        }    }    return -1;}int main(){    while(cin >> a >> b >> k)    {    memset(vis,0,sizeof(vis));    if (BFS() == -1)        cout << "impossible" << endl;    }    return 0;}
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