3 5 4

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

问题简述：就是简单的模拟倒水问题，两个杯子有六种情况：fill(1),fill(2),drop(1),drop(2),pour(1,2),pour(2,1)   用bfs分别讨论入队列即可。一开始一直impossible，没注意到在pour的时候先改变x的值后面改变y值时会有影响，记录路径用一个字符数组记         录。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;const int inf=9999999;
int vis[105][105];
int a,b,c;
int flag;struct node
{int x,y,step;char move[110][10];
};
queue<node>Q;node bfs()
{int t;node tmp,cur;cur.x = 0;cur.y = 0;cur.step = 0;vis[cur.x][cur.y] = 1;Q.push(cur);while(!Q.empty()){cur = Q.front();Q.pop();if (cur.x==c || cur.y==c){flag = 1;return cur;}//fill 1tmp = cur;tmp.x = a;if (vis[tmp.x][tmp.y]==0){vis[tmp.x][tmp.y] = 1;strcpy(tmp.move[tmp.step],"FILL(1)");tmp.step++;Q.push(tmp);} //fill 2tmp = cur;tmp.y = b;if (vis[tmp.x][tmp.y]==0){vis[tmp.x][tmp.y] = 1;strcpy(tmp.move[tmp.step],"FILL(2)");tmp.step++;Q.push(tmp);} //drop 1tmp = cur;tmp.x = 0;if (vis[tmp.x][tmp.y]==0){vis[tmp.x][tmp.y] = 1;strcpy(tmp.move[tmp.step],"DROP(1)");tmp.step++;Q.push(tmp);} //drop 2tmp = cur;tmp.y = 0;if (vis[tmp.x][tmp.y]==0){vis[tmp.x][tmp.y] = 1;strcpy(tmp.move[tmp.step],"DROP(2)");tmp.step++;Q.push(tmp);}//pour(1,2)tmp = cur;if (b-tmp.y>tmp.x){t = tmp.x;tmp.x = 0;tmp.y += t;}else if(b-tmp.y<=tmp.x){tmp.x -= (b-tmp.y);tmp.y = b;}if (vis[tmp.x][tmp.y]==0){vis[tmp.x][tmp.y] = 1;strcpy(tmp.move[tmp.step],"POUR(1,2)");tmp.step++;Q.push(tmp);  }//pour(2,1)tmp = cur;if (a-tmp.x>tmp.y){tmp.x += tmp.y;tmp.y = 0;}else if (a-tmp.x<=tmp.y){t = tmp.x;tmp.x = a;tmp.y -= (a-t);}if (vis[tmp.x][tmp.y]==0){vis[tmp.x][tmp.y] = 1;strcpy(tmp.move[tmp.step],"POUR(2,1)");tmp.step++;Q.push(tmp); }}
} int main()
{while(~scanf("%d%d%d",&a,&b,&c)){node ans;memset(vis,0,sizeof(vis));flag = 0;ans = bfs();if (flag){printf("%d\n",ans.step);for(int i=0; i<ans.step; i++)printf("%s\n",ans.move[i]);}elseprintf("impossible\n");}return 0;
}