(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\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();
}
}
}