3个水杯倒水有关问题(广度优先搜索)

3个水杯倒水问题(广度优先搜索)

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

来源

经典题目

广度优先搜索，貌似可以用A*，可惜不怎么会，以后试试。

# include <stdio.h># include <stdlib.h># include <string.h># include <string>void check (int queue[], int x, int &tail) {for (int i = 0; i <= tail; ++ i) {if (queue[i] == x) {return;}}++ tail;//printf ("%d -> ", x);queue[tail] = x;return;}int getans(int to[], int i) {if (i == to[1] * 10000 + to[2] * 100 + to[3]) return 1;return 0;}int daoshui(int queue[], int &a, int &b, int to, int from[]) {if (a) {int t = from[to] - b;if (t > a) {b += a;a = 0;return 1;}else if (t) {a -= t;b += t;return 1;}}return 0;}int main () {int n;scanf ("%d", &n);while (n --) {int from[4];int to[4];scanf ("%d %d %d %d %d %d", &from[1], &from[2], &from[3], &to[1], &to[2], &to[3]);int head = 0;int tail = 0;int step = 0; int queue[100000];queue[0] = from[1] * 10000;int flag = 0;if (queue[0] == to[1] * 10000 + to[2] * 100 + to[3]) {printf("0\n");continue;}while (head <= tail) {++ step;//printf("******************************%d************************************\n", step);int size = tail - head;for (int i = head; i <= head + size; ++ i) {int a, b, c;//printf ("|%d|", queue[i]);a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;if (daoshui(queue, a, b, 2, from)) {//printf ("(a -> b)");check(queue, a * 10000 + b * 100 + c, tail);}a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;if (daoshui(queue, a, c, 3, from)) {//printf ("(a -> c)");check(queue, a * 10000 + b * 100 + c, tail);}a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;if (daoshui(queue, b, a, 1, from)) {//printf ("(b -> a)");check(queue, a * 10000 + b * 100 + c, tail);}a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;if (daoshui(queue, b, c, 3, from)) {//printf ("(b -> c)");check(queue, a * 10000 + b * 100 + c, tail);}a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;if (daoshui(queue, c, a, 1, from)) {//printf ("(c -> a)");check(queue, a * 10000 + b * 100 + c, tail);}a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;if (daoshui(queue, c, b, 2, from)) {//printf ("(c -> b)");check(queue, a * 10000 + b * 100 + c, tail);}}//printf ("\n****************************************************************");head += size + 1;for (int i = head; i <= tail; ++ i) {if (getans(to, queue[i])) {printf ("%d\n", step);flag = 1;break;}}if (flag) break;}//printf("\n");if (!flag) printf ("-1\n");}return 0;}