非常可乐（bfs）

Problem Description 大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升　(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。   Input 三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。   Output 如果能平分的话请输出最少要倒的次数，否则输出"NO"。   Sample Input 7 4 3 4 1 3 0 0 0   Sample Output NO 3

#include <iostream>
#include <queue>
using namespace std;
int s; // 一罐可乐
int a, b; // 2个杯子
int half;


typedef struct Now {
    int s;
    int a, b;
    int step;
} Now;

queue<Now> q;
bool mark[110][110][110];


int aToB(int& a, int& aMax, int& b, int& bMax) {
    if (b == bMax) {
        return 0;
    }
    if (a == 0) {
        return 0;
    }


    // 倒了, 有改动

    // b缺口非常大，a全到进去了
    if (a <= bMax - b) {
        b += a;
        a = 0;
    } else {
        //B缺口小，a倒了一部分，满了
        a -= bMax - b;
        b = bMax;
    }
    return 1;
}


void printNow(Now& a) {
    cout << "contain: s a b:" << a.s << " " << a.a << " " << a.b << " step :" << a.step << endl;
}


bool judge(Now& n) {
    if (n.s == n.a && n.a == half || n.s == n.b && n.b == half || n.a == n.b && n.b == half) {
        return 1;
    } else {
        return 0;
    }
}


/*操作：或将目标容器倒满，或将源容器倒空。*/
int bfs(int s, int a, int b) {
    while(!q.empty()) {
        q.pop();
    }

    Now start;
    start.s = s;
    start.a = 0;
    start.b = 0;
    start.step = 0;
    mark[s][0][0] = 1;

    Now father;
    Now child;

    q.push(start);
    while(!q.empty()) {
        father = q.front();
        q.pop();
        //printNow(father);
        if(judge(father)) {
            return father.step;
        }
        int ns;
        int na;
        int nb;
        int nstep;

        //s to a
        ns = father.s;
        na = father.a;
        nb = father.b;
        nstep = father.step;

        if (aToB(ns, s, na, a) == 1) {
            if (mark[ns][na][nb] == 0) {
                //cout << "s-a" << endl;

                child.s = ns;
                child.a = na;
                child.b = nb;
                child.step = nstep + 1;
                q.push(child);
                mark[ns][na][nb] = 1;
            }
        }


        //s to b
        ns = father.s;
        na = father.a;
        nb = father.b;

        if (aToB(ns, s, nb, b) == 1) {
            if (mark[ns][na][nb] == 0) {
                //cout << "s-b" << endl;
                child.s = ns;
                child.a = na;
                child.b = nb;
                child.step = nstep + 1;
                q.push(child);
                mark[ns][na][nb] = 1;
            }
        }


        //a to s
        ns = father.s;
        na = father.a;
        nb = father.b;

        if (aToB(na, a, ns, s) == 1) {
            if (mark[ns][na][nb] == 0) {
                //cout << "a-s" << endl;
                child.s = ns;
                child.a = na;
                child.b = nb;
                child.step = nstep + 1;
                q.push(child);
                mark[ns][na][nb] = 1;
            }
        }
        //a to b
        ns = father.s;
        na = father.a;
        nb = father.b;

        if (aToB(na, a, nb, b) == 1) {
            if (mark[ns][na][nb] == 0) {
                //cout << "a-b" << endl;
                child.s = ns;
                child.a = na;
                child.b = nb;
                child.step = nstep + 1;
                q.push(child);
                mark[ns][na][nb] = 1;
            }
        }

        //b to s

        ns = father.s;
        na = father.a;
        nb = father.b;

        if (aToB(nb, b, ns, s) == 1) {
            if (mark[ns][na][nb] == 0) {
                //cout << "b-s" << endl;
                child.s = ns;
                child.a = na;
                child.b = nb;
                child.step = nstep + 1;
                q.push(child);
                mark[ns][na][nb] = 1;
            }
        }
        //b to a
        ns = father.s;
        na = father.a;
        nb = father.b;

        if (aToB(nb, b, na, a) == 1) {
            if (mark[ns][na][nb] == 0) {
                //cout << "b-a" << endl;
                child.s = ns;
                child.a = na;
                child.b = nb;
                child.step = nstep + 1;
                q.push(child);
                mark[ns][na][nb] = 1;
            }
        }


    }

    return -1;
}


int main() {


    while(cin >> s >> a >> b && s != 0 && a != 0 && b != 0) {
        half = s / 2;
        for (int i = 0; i <= s; i ++) {
            for (int j = 0; j <= a; j++)
                for (int k = 0; k <= b; k++) {
                    mark[i][j][k] = 0;
                }
        }

        if (s % 2 == 1){
            cout << "NO" << endl;
            continue;
        }

        int res = bfs(s, a, b);
        if (res == -1) {
            cout << "NO" << endl;
        } else {
            cout << res << endl;
        }
    }





}

注意：千万不要 a == b == c   //1 == C

需要 a == b && b == c

#include <iostream>
#include <queue>
#include <cmath>
using namespace std;

int sMax; //可乐的体积
int aMax,bMax; //两个杯子

int half;

typedef struct Now {
	int s;
	int a;
	int b;
	int cnt;
} Now;

bool mark[101][101][101];

queue<Now> q;

// OK
void s2a(Now& n) {
	int k = aMax - n.a;
	if (n.s >= k) {
		// 目标弄满，原来的还剩下
		n.a = aMax;
		n.s -= k;
		n.cnt++;
	} else {
		// 目标放入原来的全部，原来清空
		n.a += n.s;
		n.s = 0;
		n.cnt++;
	}
}

// OK
void s2b(Now& n) {
	int k = bMax - n.b;
	if (n.s >= k) {
		// 目标弄满，原来的还剩下
		n.b = bMax;
		n.s -= k;
		n.cnt++;
	} else {
		// 目标放入原来的全部，原来清空
		n.b += n.s;
		n.s = 0;
		n.cnt++;
	}
}


// OK
void a2s(Now& n) {
	int k = sMax - n.s;
	if (n.a >= k) {
		// 目标弄满，原来的还剩下
		n.s = sMax;
		n.a -= k;
		n.cnt++;
	} else {
		// 目标放入原来的全部，原来清空
		n.s += n.a;
		n.a = 0;
		n.cnt++;
	}
}


// OK
void a2b(Now& n) {
	int k = bMax - n.b;
	if (n.a >= k) {
		// 目标弄满，原来的还剩下
		n.b = bMax;
		n.a -= k;
		n.cnt++;
	} else {
		// 目标放入原来的全部，原来清空
		n.b += n.a;
		n.a = 0;
		n.cnt++;
	}
}


// OK
void b2s(Now& n) {
	int k = sMax - n.s;
	if (n.b >= k) {
		// 目标弄满，原来的还剩下
		n.s = sMax;
		n.b -= k;
		n.cnt++;
	} else {
		// 目标放入原来的全部，原来清空
		n.s += n.b;
		n.b = 0;
		n.cnt++;
	}
}


// OK
void b2a(Now& n) {
	int k = aMax - n.a;
	if (n.b >= k) {
		// 目标弄满，原来的还剩下
		n.a = aMax;
		n.b -= k;
		n.cnt++;
	} else {
		// 目标放入原来的全部，原来清空
		n.a += n.b;
		n.b = 0;
		n.cnt++;
	}
}

bool check(Now& n) {
	if (n.s == half && n.a == half) {
		return 1;
	} else if (n.s == half && n.b == half) {
		return 1;
	} else if (n.a == half && n.b == half) {
		return 1;
	}

	return 0;
}

int bfs() {
	while(!q.empty()) {
		q.pop();
	}


	Now chu;
	chu.s = sMax;
	chu.a = chu.b =0;
	chu.cnt = 0;
	mark[sMax][0][0] = 1;

	Now father;
	Now child;


	q.push(chu);
	while(!q.empty()) {
		father = q.front();
		q.pop();
		// 父亲出来 孩子进入
//		cout << "father: " << father.s << " " << father.a << " " << father.b << endl;

		if (check(father) == 1) {
			return father.cnt;
		}


		//S TO A
		child = father;
		s2a(child);
		if (mark[child.s][child.a][child.b] == 0) {
			mark[child.s][child.a][child.b] = 1;
			q.push(child);
		}

		//s to b
		child = father;
		s2b(child);
		if (mark[child.s][child.a][child.b] == 0) {
			mark[child.s][child.a][child.b] = 1;
			q.push(child);
		}

		//a TO s
		child = father;
		a2s(child);
		if (mark[child.s][child.a][child.b] == 0) {
			mark[child.s][child.a][child.b] = 1;
			q.push(child);
		}
		//a TO b
		child = father;
		a2b(child);
		if (mark[child.s][child.a][child.b] == 0) {
			mark[child.s][child.a][child.b] = 1;
			q.push(child);
		}

		//b TO s
		child = father;
		b2s(child);
		if (mark[child.s][child.a][child.b] == 0) {
			mark[child.s][child.a][child.b] = 1;
			q.push(child);
		}

		//b TO a
		child = father;
		b2a(child);
		if (mark[child.s][child.a][child.b] == 0) {
			mark[child.s][child.a][child.b] = 1;
			q.push(child);
		}

	}



	return  -1;

}


int main() {

	while (cin >> sMax >> aMax >> bMax) {
		if (sMax == 0 && aMax == 0 && bMax == 0) {
			return 0;
		}


		for (int i = 0; i <= 100; i++) {
			for (int j = 0; j <= 100; j++) {
				for (int k = 0; k <= 100; k++) {
					mark[i][j][k] = 0;
				}
			}
		}
		half = sMax / 2;

		if (sMax % 2 == 1) {

			cout << "NO" << endl;
			continue;
		} else {


			int res = bfs();

			if (res != -1) {
				cout << res << endl;
			} else {
				cout << "NO" << endl;
			}

		}


	}

}