NYOJ 三个水杯

时间限制：1000 ms  |  内存限制：65535 KB
难度：4
描述
给出三个水杯，大小不一，并且只有最大的水杯的水是装满的，其余两个为空杯子。三个水杯之间相互倒水，并且水杯没有标识，只能根据给出的水杯体积来计算。现在要求你写出一个程序，使其输出使初始状态到达目标状态的最少次数。
输入
第一行一个整数N(0<N<50)表示N组测试数据
接下来每组测试数据有两行，第一行给出三个整数V1 V2 V3 (V1>V2>V3 V1<100 V3>0)表示三个水杯的体积。
第二行给出三个整数E1 E2 E3 （体积小于等于相应水杯体积）表示我们需要的最终状态
输出
每行输出相应测试数据最少的倒水次数。如果达不到目标状态输出-1
样例输入
2
6 3 1
4 1 1
9 3 2
7 1 1
样例输出
3
-1
来源
经典题目
上传者

hzyqazasdf

广搜：

#include<iostream>#include<queue>#include<cstring>using namespace std;bool ever[102][102][102];struct state{int x,y,z,step;};state start,cur,goal,cup,tmp;int BFS(){queue<state>que;memset(ever,0,sizeof(ever));cur = start;que.push(cur);ever[start.x][start.y][start.z] = 1;bool success = 0;while(!que.empty()){cur = que.front();que.pop();//cout<<cur.x<<' '<<cur.y<<' '<<cur.z<<' '<<cur.step<<endl;//cout<<cup.x<<' '<<cup.y<<' '<<cup.z<<endl;if(cur.x == goal.x && cur.y == goal.y && cur.z == goal.z){goal.step = cur.step;success = 1;break;}if(cur.x  != 0 && cur.y < cup.y ){if(cur.x >= cup.y - cur.y)     tmp.x = cur.x - (cup.y - cur.y);else tmp.x = 0;tmp.y = cur.y + cur.x - tmp.x;tmp.z = cur.z;tmp.step = cur.step + 1;if(!ever[tmp.x][tmp.y][tmp.z]){ever[tmp.x][tmp.y][tmp.z] = 1;que.push(tmp);}}if(cur.x  != 0 && cur.z < cup.z ){if(cur.x >= cup.z - cur.z)     tmp.x = cur.x - (cup.z - cur.z);else tmp.x = 0;tmp.z = cur.z + cur.x - tmp.x;tmp.y = cur.y;tmp.step = cur.step + 1;if(!ever[tmp.x][tmp.y][tmp.z]){ever[tmp.x][tmp.y][tmp.z] = 1;que.push(tmp);}}if(cur.y  != 0 && cur.x < cup.x ){if(cur.y >= cup.x - cur.x)     tmp.y = cur.y - (cup.x - cur.x);else tmp.y = 0;tmp.x = cur.x + cur.y - tmp.y;tmp.z = cur.z;tmp.step = cur.step + 1;if(!ever[tmp.x][tmp.y][tmp.z]){ever[tmp.x][tmp.y][tmp.z] = 1;que.push(tmp);}}if(cur.y  != 0 && cur.z < cup.z ){if(cur.y >= cup.z - cur.z)     tmp.y = cur.y - (cup.z - cur.z);else tmp.y = 0;tmp.z = cur.z + cur.y - tmp.y;tmp.x = cur.x;tmp.step = cur.step + 1;if(!ever[tmp.x][tmp.y][tmp.z]){ever[tmp.x][tmp.y][tmp.z] = 1;que.push(tmp);}}if(cur.z  != 0 && cur.y < cup.y ){if(cur.z >= cup.y - cur.y)     tmp.z = cur.z - (cup.y - cur.y);else tmp.z = 0;tmp.y = cur.y + cur.z - tmp.z;tmp.x = cur.x;tmp.step = cur.step + 1;if(!ever[tmp.x][tmp.y][tmp.z]){ever[tmp.x][tmp.y][tmp.z] = 1;que.push(tmp);}}if(cur.z  != 0 && cur.x < cup.x ){if(cur.z >= cup.x - cur.x)     tmp.z = cur.z - (cup.x - cur.x);else tmp.z = 0;tmp.x = cur.x + cur.z - tmp.z;tmp.y = cur.y;tmp.step = cur.step + 1;if(!ever[tmp.x][tmp.y][tmp.z]){ever[tmp.x][tmp.y][tmp.z] = 1;que.push(tmp);}}}if(success)return goal.step;else return -1;}int main(){int T;cin>>T;while(T--){cin>>cup.x>>cup.y>>cup.z;cin>>goal.x>>goal.y>>goal.z;start.x = cup.x;start.y = 0;start.z = 0;start.step = 0;cout<<BFS()<<endl;}return 0;}