三个水杯

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

26 3 14 1 19 3 27 1 1

样例输出

3-1

 #include <stdio.h>#include <string.h>#include <iostream>#include <queue>#define Max_w 205using namespace std;int v1_e, v1_h;int v2_e, v2_h;int v3_e, v3_h;int visited[Max_w][Max_w][Max_w] = {0};typedef struct cup{int v1;int v2;int v3;int step;}Cup;void bfs(Cup c);int main (){int N;int i, j, d;Cup c;scanf("%d", &N);while(N--){memset(visited, 0, sizeof(visited));scanf("%d %d %d", &v1_h, &v2_h, &v3_h);scanf("%d %d %d", &v1_e, &v2_e, &v3_e);if(v1_h == v1_e && 0 == v2_e && 0 == v3_e){             printf("0\n");             continue;        }visited[v1_h][0][0] = 1;c.v1 = v1_h; c.v2 = 0; c.v3 = 0;c.step = 0;bfs(c);}return 0;}void bfs(Cup c){int i;queue<Cup> Q;Cup temp, temp1;Q.push(c);while (!Q.empty()){temp1 = Q.front();Q.pop();if(temp1.v1 == v1_e && temp1.v2 == v2_e && temp1.v3 == v3_e){printf("%d\n", temp1.step);return ;}for(i = 0; i < 6; i++){temp = temp1;if(i == 0 && temp.v1 != 0 && temp.v2 != v2_h){if(temp.v1 >= v2_h - temp.v2){temp.v1 -= (v2_h - temp.v2);temp.v2 = v2_h;}else{temp.v2 += temp.v1;temp.v1 = 0;}temp.step ++;if(!visited[temp.v1][temp.v2][temp.v3]){visited[temp.v1][temp.v2][temp.v3] = 1;Q.push(temp);}}if(i == 1 && temp.v1 != 0 && temp.v3 != v3_h){if(temp.v1 >= v3_h - temp.v3){temp.v1 -= (v3_h - temp.v3);temp.v3 = v3_h;}else{temp.v3 += temp.v1;temp.v1 = 0;}temp.step ++;if(!visited[temp.v1][temp.v2][temp.v3]){visited[temp.v1][temp.v2][temp.v3] = 1;Q.push(temp);}}if(i ==  2 && temp.v2 != 0 && temp.v3 != v3_h){if(temp.v2 >= v3_h - temp.v3){temp.v2 -= (v3_h - temp.v3);temp.v3 = v3_h;}else{temp.v3 += temp.v2;temp.v2 = 0;}temp.step ++;if(!visited[temp.v1][temp.v2][temp.v3]){visited[temp.v1][temp.v2][temp.v3] = 1;Q.push(temp);}}if(i == 3 && temp.v2 != 0 && temp.v1 != v1_h){if(temp.v2 >= v1_h - temp.v1){temp.v2 -= (v1_h - temp.v1);temp.v1 = v1_h;}else{temp.v1 += temp.v2;temp.v2 = 0;}temp.step ++;if(!visited[temp.v1][temp.v2][temp.v3]){visited[temp.v1][temp.v2][temp.v3] = 1;Q.push(temp);}}if(i == 4 && temp.v3 != 0 && temp.v2 != v2_h){if(temp.v3 >= v2_h - temp.v2){temp.v3 -= (v2_h - temp.v2);temp.v2 = v2_h;}else{temp.v2 += temp.v3;temp.v3 = 0;}temp.step ++;if(!visited[temp.v1][temp.v2][temp.v3]){visited[temp.v1][temp.v2][temp.v3] = 1;Q.push(temp);}}if(i == 5 && temp.v3 != 0 && temp.v1 != v1_h){if(temp.v3 >= v1_h - temp.v1){temp.v3 -= (v1_h - temp.v1);temp.v1 = v1_h;}else{temp.v1 += temp.v3;temp.v3 = 0;}temp.step ++;if(!visited[temp.v1][temp.v2][temp.v3]){visited[temp.v1][temp.v2][temp.v3] = 1;Q.push(temp);}}}}printf("%d\n", -1);}