(step4.2.5)hdu 1495(非常可乐——BFS)

题目大意：输入三个整数 a,b,c.   a : 可乐瓶的容量，b: 甲杯的容量 ，c: 乙杯的容量。问能否用这三个被来实现饮料的平分？？？如果可以输出倒饮料的次数,

否则输出NO

解题思路:BFS

1）本题的考点其实在于将标记数组由二维数组变为三维数组。遍历状态由使用for()循环变为手动枚举，一个一个的if()

代码如下:

/*
 * 1495_2.cpp
 *
 *  Created on: 2013年8月16日
 *      Author: Administrator
 */

#include <iostream>
#include <queue>

using namespace std;
const int maxn = 102;
bool visited[maxn][maxn][maxn];

int a, b, c;
struct State {
	int a;
	int b;
	int c;
	int v;
};

bool checkState(State st) {
	if (!visited[st.a][st.b][st.c]) {
		return true;
	}

	return false;
}

void bfs() {
	queue<State> q;
	State st, now, next;

	st.a = a;
	st.b = 0;
	st.c = 0;
	st.v = 0;
	q.push(st);
	memset(visited,0,sizeof(visited) );
	visited[st.a][st.b][st.c] = 1;
	while (!q.empty()) {
		now = q.front();

		//有2个等于a/2就结束
		if ((now.a == a / 2 && now.b == a / 2)
				|| (now.a == a / 2 && now.c == a / 2)
				|| (now.c == a / 2 && now.b == a / 2)) {
			printf("%d
", now.v);
			return ;
		}

		/**
		 * 若a杯中的饮料的体积不为0,
		 * 则枚举出将a杯中的饮料倒到其他杯中....
		 */
		if (now.a != 0) {
			/**
			 * 关键举得理解:now.a > b - now.b
			 * now.a : now状态下a杯中的饮料的体积
			 * b : b杯的体积
			 * now.b :now状态下b杯中的饮料的体积
			 *
			 */
			if (now.a > b - now.b) {//now.a > b - now.b。且倒不完
				next.a = now.a - (b - now.b);
				next.b = b;
				next.c = now.c;
				next.v = now.v + 1;
			} else {//倒完了
				next.a = 0;
				next.b = now.b + now.a;
				next.c = now.c;
				next.v = now.v + 1;
			}

			if (checkState(next)) {
				q.push(next);
				visited[next.a][next.b][next.c] = 1;
			}

			if (now.a > c - now.c) {
				next.a = now.a - (c - now.c);
				next.b = now.b;
				next.c = c;
				next.v = now.v + 1;
			} else {
				next.a = 0;
				next.b = now.b;
				next.c = now.c + now.a;
				next.v = now.v + 1;
			}

			if (checkState(next)) {
				q.push(next);
				visited[next.a][next.b][next.c] = 1;
			}
		}

		if (now.b != 0) {
			if (now.b > a - now.a) {
				next.a = a;
				next.b = now.b - (a - now.a);
				next.c = now.c;
				next.v = now.v + 1;
			} else {
				next.a = now.a + now.b;
				next.b = 0;
				next.c = now.c;
				next.v = now.v + 1;
			}

			if (checkState(next)) {
				q.push(next);
				visited[next.a][next.b][next.c] = 1;
			}

			if (now.b > c - now.c) {
				next.a = now.a ;
				next.b = now.b - (c - now.c);
				next.c = c;
				next.v = now.v + 1;
			} else {
				next.a = now.a;
				next.b = 0;
				next.c = now.c + now.b;
				next.v = now.v + 1;
			}

			if (checkState(next)) {
				q.push(next);
				visited[next.a][next.b][next.c] = 1;
			}
		}

		if (now.c != 0) {
			if (now.c > b - now.b) {
				next.a = now.a ;
				next.b = b;
				next.c = now.c - (b - now.b);
				next.v = now.v + 1;
			} else {
				next.a = now.a;
				next.b = now.b + now.c;
				next.c = 0;
				next.v = now.v + 1;
			}

			if (checkState(next)) {
				q.push(next);
				visited[next.a][next.b][next.c] = 1;
			}

			if (now.c > a - now.a) {
				next.a = a;
				next.b = now.b;
				next.c = now.c - (a - now.a);
				next.v = now.v + 1;
			} else {
				next.a = now.a + now.c;
				next.b = now.b;
				next.c = 0;
				next.v = now.v + 1;
			}

			if (checkState(next)) {
				q.push(next);
				visited[next.a][next.b][next.c] = 1;
			}
		}
		q.pop();
	}

	printf("NO
");
}
int main() {

	while(scanf("%d%d%d",&a,&b,&c)!=EOF,a+b+c){
		if(a%2 == 1){
			printf("NO
");
		}else{
			bfs();
		}
	}
}