2021-bfsTRN-J

https://vjudge.net/contest/425322#problem/J

题目概括

给定容积的三个杯子倒可乐（必须倒到一个杯子没有可乐或者一个杯子装满），问能否平分。

思路

很像数学题（他确实是），为什么和bfs有关呢？
因为三个杯子的可乐体积代表了一个状态。状态数有限，所以可以bfs。每次拓展时，分六种情况讨论是那个杯子里的可乐倒到那个杯子里面去，这题就搞定了。
（代码好长）
（数学做法不会，有空学~~）

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>

using namespace std;
int a,b,c,ans;
int vis[105][105][105];
struct node
{
    int a,b,c,step;
};
int bfs()
{
    queue<node>q;
    node now;
    now.a=a,now.b=0,now.c=0,now.step=0;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if((now.a==a/2&&now.b==a/2)||(now.a==a/2&&now.c==a/2)||(now.b==a/2&&now.c==a/2))
        {
            while(!q.empty()) q.pop();
            return now.step;
        }
        node next;
        if(now.a>0)
        {
            if(now.a+now.b<=b)
            {
                next.a=0,next.b=now.a+now.b,next.c=now.c,next.step=now.step+1;
            }
            else
            {
                next.a=now.a-(b-now.b),next.b=b,next.c=now.c,next.step=now.step+1;
            }
            if(!vis[next.a][next.b][next.c])
            {
                vis[next.a][next.b][next.c]=1;
                q.push(next);
            }
        }
        if(now.a>0)
        {
            if(now.a+now.c<=c)
            {
                next.a=0,next.b=now.b,next.c=now.a+now.c,next.step=now.step+1;
            }
            else
            {
                next.a=now.a-(c-now.c),next.b=now.b,next.c=c,next.step=now.step+1;
            }
            if(!vis[next.a][next.b][next.c])
            {
                vis[next.a][next.b][next.c]=1;
                q.push(next);
            }
        }
        if(now.b>0)
        {
            if(now.a+now.b<=a)
            {
                next.a=now.a+now.b,next.b=0,next.c=now.c,next.step=now.step+1;
            }
            else
            {
                next.a=a,next.b=now.b-(a-now.a),next.c=now.c,next.step=now.step+1;
            }
            if(!vis[next.a][next.b][next.c])
            {
                vis[next.a][next.b][next.c]=1;
                q.push(next);
            }
        }
        if(now.b>0)
        {
            if(now.b+now.c<=c)
            {
                next.a=now.a,next.b=0,next.c=now.b+now.c,next.step=now.step+1;
            }
            else
            {
                next.a=now.a,next.b=now.b-(c-now.c),next.c=c,next.step=now.step+1;
            }
            if(!vis[next.a][next.b][next.c])
            {
                vis[next.a][next.b][next.c]=1;
                q.push(next);
            }
        }
        if(now.c>0)
        {
            if(now.c+now.a<=a)
            {
                next.a=now.c+now.a,next.b=now.b,next.c=0,next.step=now.step+1;
            }
            else
            {
                next.a=a,next.b=now.b,next.c=now.c-(a-now.a),next.step=now.step+1;
            }
            if(!vis[next.a][next.b][next.c])
            {
                vis[next.a][next.b][next.c]=1;
                q.push(next);
            }
        }
        if(now.c>0)
        {
            if(now.b+now.c<=b)
            {
                next.a=now.a,next.b=now.b+now.c,next.c=0,next.step=now.step+1;
            }
            else
            {
                next.a=now.a,next.b=b,next.c=now.c-(b-now.b),next.step=now.step+1;
            }
            if(!vis[next.a][next.b][next.c])
            {
                vis[next.a][next.b][next.c]=1;
                q.push(next);
            }
        }

    }
    return -1;
}
int main()
{
    while(~scanf("%d%d%d",&a,&b,&c))
    {
        if(!a&&!b&&!c) break;
        if(a%2)
        {
            printf("NO\n");continue;
        }
        ans=0;
        memset(vis,0,sizeof(vis));
        vis[a][0][0]=1;
        int ans=bfs();
        if(ans==-1) printf("NO\n");
        else printf("%d\n",ans);
    }
    return 0;
}