非常可乐

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17457 Accepted Submission(s): 7092

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
 

Author

seeyou

vis[][]数组表示a,b瓶中可乐是否存在,分6种情况(s->a,s->b,a->b,b->a,a->s,b->s),进行bfs,当a,b到s的一半时输出步数

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <cmath>

using namespace std;

int n;
int m;
int s;

int vis[105][105];

typedef struct node {
    int s;
    int a;
    int b;
    int t;
} node;

int bfs() {
    queue<node> q;
    node now;
    node to;
    memset(vis,0,sizeof(vis));
    now.s = s;
    now.a = 0;
    now.b = 0;
    now.t = 0;
    q.push(now);
    vis[now.a][now.b] = 1;


    while(!q.empty()) {
        now = q.front();
        q.pop();

        if(now.s == s/2 && now.a == s/2) {
            return now.t;
        }
        if(now.s && now.a != n) { //s->a
            int c = n - now.a;
            if(now.s >= c) {
                to.a = n;
                to.s = now.s - c;
            } else {
                to.a = now.a + now.s;
                to.s = 0;
            }
            to.b = now.b;
            to.t = now.t + 1;
            if(!vis[to.a][to.b]) {
                q.push(to);
                vis[to.a][to.b] = 1;
            }
        }
        if(now.s && now.b != m) { //s->b
            int c = m - now.b;
            if(now.s >= c) {
                to.b = m;
                to.s = now.s - c;
            } else {
                to.b = now.b + now.s;
                to.s = 0;
            }
            to.a = now.a;
            to.t = now.t + 1;
            if(!vis[to.a][to.b]) {
                q.push(to);
                vis[to.a][to.b] = 1;
            }
        }
        if(now.a && now.s != s) { //a->s
            int c = s - now.s;
            if(now.a >= c) {
                to.s = s;
                to.a = now.a - c;
            } else {
                to.s = now.a + now.s;
                to.a = 0;
            }
            to.b = now.b;
            to.t = now.t + 1;
            if(!vis[to.a][to.b]) {
                q.push(to);
                vis[to.a][to.b] = 1;
            }
        }
        if(now.a && now.b != m) { //a->b
            int c = m - now.b;
            if(now.a >= c) {
                to.b = m;
                to.a = now.a - c;
            } else  {
                to.b = now.b + now.a;
                to.a = 0;
            }
            to.s = now.s;
            to.t = now.t + 1;
            if(!vis[to.a][to.b]) {
                q.push(to);
                vis[to.a][to.b] = 1;
            }
        }

        if(now.b && now.a != n) {//b->a
            int c = n - now.a;
            if(now.b >= c) {
                to.a = n;
                to.b = now.b - c;
            } else {
                to.a = now.b + now.a;
                to.b = 0;
            }
            to.s = now.s;
            to.t = now.t + 1;
            if(!vis[to.a][to.b]) {
                q.push(to);
                vis[to.a][to.b]=1;
            }
        }
        if(now.b && now.s != s) {//b->s
            int c = s - now.s;
            if(now.b >= c) {
                to.s = s;
                to.b = now.b - c;
            } else {
                to.s = now.s + now.b;
                to.b = 0;
            }
            to.a = now.a;
            to.t = now.t + 1;
            if(!vis[to.a][to.b]) {
                q.push(to);
                vis[to.a][to.b] = 1;
            }
        }
    }

    return 0;
}


int main(){
    while(cin >> s >> n >> m) {
        if(!s && !n && !m) {
            break;
        }
        if(s % 2) {
            cout << "NO" << endl;
        }
        if(m > n) {
            swap(m,n);
        }
        int res = bfs();
        if(res) {
            cout << res << endl;
        } else {
            cout << "NO" << endl;
        }
    }

    return 0;
}