poj3414 Pots

题意：有两个无刻度的容量分别为A,B升的杯子，通过一些操作使某一个杯子中有C升的水。

1. FILL(i) ，将i这个杯子中的水接满
2. DROP(i)，将i这个杯子中的水倒掉
3. POUR(i,j)，将i这个杯子中的水倒入j这个杯子，能倒完就倒完，倒不完就留在杯子中。

问达到目标状态的操作次数最少的方案是什么？

思路：BFS+路径输出。共六种操作FILL(1)，FILL(2)，DROP(1)，DROP(2)，POUR(1,2)，POUR(2,1)。路径另外去开一个数组去存来的方向就好了。如3 4这状态是由0 4来的，path[3][4]里面存0 4。

//#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;const int MAXN = 105;int A, B, C;struct state//状态{    int a, b, step;}NOW, NEXT;struct PATH//路径{    int a, b, op;}path[MAXN][MAXN];void putout(int x,int y)//路径输出{    if (x == 0 && y == 0) return ;    putout(path[x][y].a, path[x][y].b);    if (path[x][y].op == 1) printf("FILL(1)\n");    else if (path[x][y].op == 2) printf("FILL(2)\n");    else if (path[x][y].op == 3) printf("DROP(1)\n");    else if (path[x][y].op == 4) printf("DROP(2)\n");    else if (path[x][y].op == 5) printf("POUR(1,2)\n");    else if (path[x][y].op == 6) printf("POUR(2,1)\n");}bool vis[MAXN][MAXN];bool bfs(){    memset(vis, 0, sizeof(vis));    NOW.a = 0, NOW.b = 0, NOW.step = 0; vis[0][0] = true;    queue<state> q;    q.push(NOW);    while (!q.empty())    {        NOW = q.front();        q.pop();        if (NOW.a == C || NOW.b == C)//输出答案        {            printf("%d\n",NOW.step);            putout(NOW.a, NOW.b);            return true;        }//FILL(i)        //FILL(1)        NEXT.a = A; NEXT.b = NOW.b; NEXT.step = NOW.step + 1;        if (!vis[NEXT.a][NEXT.b])        {            vis[NEXT.a][NEXT.b] = true;            q.push(NEXT);            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//记录路径            path[NEXT.a][NEXT.b].op = 1;//记录操作        }        //FILL(2)        NEXT.a = NOW.a; NEXT.b = B; NEXT.step = NOW.step + 1;        if (!vis[NEXT.a][NEXT.b])        {            vis[NEXT.a][NEXT.b] = true;            q.push(NEXT);            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//记录路径            path[NEXT.a][NEXT.b].op = 2;//记录操作        }//DROP(i)        //DROP(1)        NEXT.a = 0; NEXT.b = NOW.b; NEXT.step = NOW.step + 1;        if (!vis[NEXT.a][NEXT.b])        {            vis[NEXT.a][NEXT.b] = true;            q.push(NEXT);            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//记录路径            path[NEXT.a][NEXT.b].op = 3;//记录操作        }        //DROP(2)        NEXT.a = NOW.a; NEXT.b = 0; NEXT.step = NOW.step + 1;        if (!vis[NEXT.a][NEXT.b])        {            vis[NEXT.a][NEXT.b] = true;            q.push(NEXT);            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//记录路径            path[NEXT.a][NEXT.b].op = 4;//记录操作        }//POUR(i,j)        //POUR(1,2)        if (NOW.a >= B - NOW.b)//i能把j倒满        {            NEXT.a = NOW.a - (B - NOW.b); NEXT.b = B; NEXT.step = NOW.step + 1;        }        else //倒不满j        {            NEXT.a = 0; NEXT.b = NOW.b + NOW.a; NEXT.step = NOW.step + 1;        }        if (!vis[NEXT.a][NEXT.b])        {            vis[NEXT.a][NEXT.b] = true;            q.push(NEXT);            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//记录路径            path[NEXT.a][NEXT.b].op = 5;//记录操作        }        //POUR(2,1)        if (NOW.b >= A - NOW.a)//j能把i倒满        {            NEXT.a = A; NEXT.b = NOW.b - (A - NOW.a); NEXT.step = NOW.step + 1;        }        else //倒不满i        {            NEXT.a = NOW.a + NOW.b; NEXT.b = 0; NEXT.step = NOW.step + 1;        }        if (!vis[NEXT.a][NEXT.b])        {            vis[NEXT.a][NEXT.b] = true;            q.push(NEXT);            path[NEXT.a][NEXT.b].a = NOW.a; path[NEXT.a][NEXT.b].b = NOW.b;//记录路径            path[NEXT.a][NEXT.b].op = 6;//记录操作        }    }    return false;}int main(){    while (~scanf("%d%d%d", &A, &B, &C))    {        if (bfs() == false) printf("impossible\n");    }    return 0;}/*3 4 5*//*第一个样例解释:3 5 4         i  jFILL(2)       0  5POUR(2,1)     3  2DROP(1)       0  2POUR(2,1)     2  0FILL(2)       2  5POUR(2,1)     3  4*/