HDU - 1495 非常可樂(倒水問題BFS)

Description

大家一定覺的運動以後喝可樂是一件很愜意的事情，但是seeyou卻不這麼認爲。因爲每次當seeyou買了可樂以後，阿牛就要求和seeyou一起分享這一瓶可樂，而且一定要喝的和seeyou一樣多。但seeyou的手中只有兩個杯子，它們的容量分別是N 毫升和M 毫升 可樂的體積爲S （S<101）毫升　(正好裝滿一瓶) ，它們三個之間可以相互倒可樂 (都是沒有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聰明的ACMER你們說他們能平分嗎？如果能請輸出倒可樂的最少的次數，如果不能輸出"NO"。

 

Input

三個整數 : S 可樂的體積 , N 和 M是兩個杯子的容量，以"0 0 0"結束。

 

Output

如果能平分的話請輸出最少要倒的次數，否則輸出"NO"。

 

Sample Input

7 4 3 4 1 3 0 0 0

 

Sample Output

NO 3

題解：先設計M<N<S，就只有6個方向

S -> N, S->M, N->M,M->N,M->S,N->S;

利用這六個方向衍生出去，一直判斷到N和S杯相等與N == S/2，否者就不能分均

<pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN = 101;
int S,N,M;
bool vis[MAXN][MAXN];
struct node{
    int x,y,all,step;
};
//M<N<S
void BFS(){
    queue<node> Q;
    node temp1,temp2;
    temp1.x = 0,temp1.y = 0,temp1.all = S,temp1.step = 0;
    Q.push(temp1);
    vis[temp1.x][temp1.y] = true;
    while(!Q.empty()){
        temp1 = Q.front();Q.pop();
        if(temp1.y == S/2 && temp1.y == temp1.all){
            printf("%d\n",temp1.step);return;
        }
        //S->M
        if(temp1.all + temp1.x > M){
            temp2.x = M, temp2.all=temp1.all+temp1.x-M;
            temp2.y = temp1.y,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }else{
            temp2.x = temp1.all+temp1.x,temp2.y = temp1.y;
            temp2.all = 0,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }
        //S->N
        if(temp1.all + temp1.y > N){
            temp2.x = temp1.x, temp2.all = temp1.all+temp1.y-N;
            temp2.y = N, temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }else{
            temp2.x = temp1.x, temp2.all = 0;
            temp2.y = temp1.all+temp1.y,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }
        //M->N
        if(temp1.x+temp1.y > N){
            temp2.all = temp1.all,temp2.y = N;
            temp2.x = temp1.x+temp1.y-N,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }else{
            temp2.all = temp1.all,temp2.y = temp1.x+temp1.y;
            temp2.x = 0,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }
        //N->M
        if(temp1.x+temp1.y > M){
            temp2.all = temp1.all,temp2.x = M;
            temp2.y = temp1.y+temp1.x-M,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }else{
            temp2.all = temp1.all,temp2.x = temp1.x+temp1.y;
            temp2.y = 0,temp2.step = temp1.step+1;
            if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        }
        //N->S
        temp2.all = temp1.y+temp1.all,temp2.x = temp1.x;
        temp2.y = 0,temp2.step = temp1.step + 1;
        if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
        //M->S
        temp2.all = temp1.x+temp1.all,temp2.y = temp1.y;
        temp2.x = 0,temp2.step = temp1.step + 1;
        if(!vis[temp2.x][temp2.y]){Q.push(temp2);vis[temp2.x][temp2.y] = true;}
    }
    printf("NO\n");
}

int main(){
    while(scanf("%d%d%d",&S,&N,&M)==3 && S&&N&&M){
        memset(vis,false,sizeof(vis));
        if(S%2)printf("NO\n");
        else{if(M > N)swap(M,N);BFS();}
    }
    return 0;
}