(step4.2.5)hdu 1495(非常可乐——BFS)

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本文介绍了一个使用广度优先搜索(BFS)解决饮料平分问题的算法。问题要求给定三个容器的容量，判断是否能通过倒饮料的方式使两个容器中的饮料量相等并达到总量的一半。代码详细展示了状态转移的过程，并通过三维数组记录已访问的状态。

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\n", 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\n");
}
int main() {

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