(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();}}}