非常可乐(BFS)

非常可乐

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17358    Accepted Submission(s): 7042

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 广度优先搜索,六种情况,判断好往哪里倒,倒的时候又有哪些情况就可以,最后如果两个相等,另一个是零,才是平分,这里判断改了好多次。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int n,m,s;//代表了三个瓶子的容量
int vis[120][120][120];//标记三个瓶子的状态是否出现过
struct node{
    int n,m,s;//记录每一次瓶子中可乐的状态
    int step;//记录倒可乐的次数
}rounds,nextrounds;
void Bfs(){//用广度优先搜索
    queue<node>q;
    int i;
    memset(vis,0,sizeof(vis));
    q.push(rounds);
    vis[rounds.s][rounds.n][rounds.m] = 1;
    while(!q.empty()){
        rounds = q.front();
        //printf("s:%d n:%d m:%d\n",rounds.s,rounds.n,rounds.m);
        if((rounds.s==rounds.n&&rounds.m==0)||(rounds.s==rounds.m&&rounds.n==0)||(rounds.m==rounds.n&&rounds.s==0)){//平分,输出次数,返回,注意相等时不能为0,否则一开始0,0程序就结束了
            printf("%d\n",rounds.step);//上面的判断改了很多次,一是一开始两个杯子都是0,是相等的但是不行,所以相等但不为零,二是及时相等相加必须为总数,否则样例1就不会输出no了
            return;
        }
        q.pop();
        for(i = 0; i < 6; i++){//6中情况s->n,n->s,s->m,m->s,n->m,m->n;重点
            if(i==0){//s->n
                if((n-rounds.n)>=rounds.s){//如果n中所剩空间大于等于s中的可乐,就可以把s中的可乐全部倒在n中
                    nextrounds.n = rounds.n + rounds.s;//原有的加上s中的
                    nextrounds.s = 0;//全部倒在n中了
                    nextrounds.m = rounds.m;//另一个瓶子不变
                    nextrounds.step = rounds.step + 1;//次数加一
                }
                else{//如果n所剩空间小于s中的可乐量,则把n倒满,s还剩下一部分
                    nextrounds.n = n;//n倒满了
                    nextrounds.s = rounds.s-(n-rounds.n);//s中剩下部分为原来s减去n中没装满的那一部分
                    nextrounds.m = rounds.m;//不变
                    nextrounds.step = rounds.step + 1;//次数加1
                }
                if(vis[nextrounds.s][nextrounds.n][nextrounds.m]==0){
                    q.push(nextrounds);
                    vis[nextrounds.s][nextrounds.n][nextrounds.m] = 1;
                }
            }
            else if(i==1){//n->s
                if((s-rounds.s)>=rounds.n){//s中所剩空间大于n的可乐量,可以把n中倒完
                    nextrounds.s = rounds.s + rounds.n;
                    nextrounds.n = 0;
                    nextrounds.m = rounds.m;
                    nextrounds.step = rounds.step + 1;
                }
                else{//s中所剩空间小于n中可乐量,可以把s倒满,n中剩余
                    nextrounds.s = s;
                    nextrounds.n = rounds.n - (s-rounds.s);
                    nextrounds.m = rounds.m;
                    nextrounds.step = rounds.step + 1;
                }
                if(vis[nextrounds.s][nextrounds.n][nextrounds.m]==0){
                    q.push(nextrounds);
                    vis[nextrounds.s][nextrounds.n][nextrounds.m] = 1;
                }
            }
            else if(i==2){//s->m
                if((m-rounds.m)>=rounds.s){//m中剩余空间大于s可乐量,s倒完
                    nextrounds.m = rounds.m + rounds.s;
                    nextrounds.s = 0;
                    nextrounds.n = rounds.n;
                    nextrounds.step = rounds.step + 1;
                }
                else{//m剩余空间小于s可乐量,m倒满,s剩余
                    nextrounds.m = m;
                    nextrounds.s = rounds.s - (m-rounds.m);
                    nextrounds.n = rounds.n;
                    nextrounds.step = rounds.step + 1;
                }
                if(vis[nextrounds.s][nextrounds.n][nextrounds.m]==0){
                    q.push(nextrounds);
                    vis[nextrounds.s][nextrounds.n][nextrounds.m] = 1;
                }
            }
            else if(i==3){//m->s
                    if((s-rounds.s)>=rounds.m){//s中剩余空间大于m可乐量,m倒完
                    nextrounds.s = rounds.m + rounds.s;
                    nextrounds.m = 0;
                    nextrounds.n = rounds.n;
                    nextrounds.step = rounds.step + 1;
                }
                else{//s剩余空间小于m可乐量,s倒满,m剩余
                    nextrounds.s = s;
                    nextrounds.m = rounds.m - (s-rounds.s);
                    nextrounds.n = rounds.n;
                    nextrounds.step = rounds.step + 1;
                }
                if(vis[nextrounds.s][nextrounds.n][nextrounds.m]==0){
                    q.push(nextrounds);
                    vis[nextrounds.s][nextrounds.n][nextrounds.m] = 1;
                }
            }
            else if(i==4){//n->m;
                    if((m-rounds.m)>=rounds.n){//m中剩余空间大于n可乐量,n倒完
                    nextrounds.m = rounds.m + rounds.n;
                    nextrounds.n = 0;
                    nextrounds.s = rounds.s;
                    nextrounds.step = rounds.step + 1;
                }
                else{//m剩余空间小于n可乐量,m倒满,n剩余
                    nextrounds.m = m;
                    nextrounds.n = rounds.n - (m-rounds.m);
                    nextrounds.s = rounds.s;
                    nextrounds.step = rounds.step + 1;
                }
                if(vis[nextrounds.s][nextrounds.n][nextrounds.m]==0){
                    q.push(nextrounds);
                    vis[nextrounds.s][nextrounds.n][nextrounds.m] = 1;
                }
            }
            else{//m->n
                    if((n-rounds.n)>=rounds.m){//n中剩余空间大于m可乐量,m倒完
                    nextrounds.n = rounds.n + rounds.m;
                    nextrounds.m = 0;
                    nextrounds.s = rounds.s;
                    nextrounds.step = rounds.step + 1;
                }
                else{//n剩余空间小于m可乐量,n倒满,m剩余
                    nextrounds.n = n;
                    nextrounds.m = rounds.m - (n-rounds.n);
                    nextrounds.s = rounds.s;
                    nextrounds.step = rounds.step + 1;
                }
                if(vis[nextrounds.s][nextrounds.n][nextrounds.m]==0){
                    q.push(nextrounds);
                    vis[nextrounds.s][nextrounds.n][nextrounds.m] = 1;
                }
            }
        }
    }
    printf("NO\n");//如果队列为空了说明方法都用完了,不可能平分
    return;
}
int main(){
    int i;
    while(~scanf("%d%d%d",&s,&n,&m)){
        if(s==0&&n==0&&m==0)break;
        rounds.s = s;
        rounds.n = 0;
        rounds.m = 0;
        rounds.step = 0;
        if(s%2){//奇数肯定不能平分
            printf("NO\n");
        }
        else{
            Bfs();
        }
    }
    return 0;
}
    原文作者:BFS
    原文地址: https://blog.csdn.net/codeswarrior/article/details/78387455
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞