hdu 1485 非常可樂 題解

參見小白書130頁-131頁，關鍵是在狀態的存儲，一開始用L*100+M*10+S存儲每一個狀態,L,M,S分別表示第一個第二個第三個杯子所含的可樂的量，不能過，因爲這樣的編碼有重複。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 105;
typedef int state[3];
int L,M,S;
state queues[10500];
int vis[N][N][N];
int dis[N][N][N];
int fronts,rear;
void cal(int &l,int &m,int &s,int n)
{
    int all;
    switch(n)
    {
        case 0:
            all = M - m;
            if(all>l){m += l;l = 0;}
            else {l-=all;m = M;}break;
        case 1:
            all = S - s;
            if(all>l){s += l;l = 0;}
            else {l-=all;s = S;}break;
        case 2:
            all = L - l;
            if(all>m){l += m;m = 0;}
            else {m-=all;l = L;}break;
        case 3:
            all = S - s;
            if(all>m){s += m;m = 0;}
            else {m-=all;s = S;}break;
        case 4:
            all = L - l;
            if(all>s){l += s;s = 0;}
            else {s-=all;l = L;}break;
        case 5:
            all = M - m;
            if(all>s){m += s;s = 0;}
            else {s-=all;m = M;}break;
    }
}
bool chack(state u)
{
    int l = u[0],m = u[1],s = u[2];
    if(l == m && l+m == L)return 1;
    else if(l == s && l+s == L)return 1;
    else if(m == s && m+s == L)return 1;
    else return 0;
}
int bfs()
{
    fronts = 0;rear = 1;
    queues[fronts][0] = L;queues[fronts][1] = queues[fronts][2] = 0;
    vis[L][0][0] = 1;
    dis[L][0][0] = 0;
    while(rear > fronts){
        state& u = queues[fronts++];
        if(chack(u))return dis[u[0]][u[1]][u[2]];
        int l = u[0];
        int m = u[1];
        int s = u[2];
        for(int i = 0;i<6;i++){
            int newl = l,newm = m,news = s;
            cal(newl,newm,news,i);
            if(!vis[newl][newm][news]){
                queues[rear][0] = newl;queues[rear][1] = newm;queues[rear][2] = news;rear++;
                vis[newl][newm][news] = 1;
                dis[newl][newm][news] = dis[l][m][s] + 1;
            }
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d%d%d",&L,&M,&S) == 3){
        if(L == 0 && M == 0 && S == 0)break;
        memset(vis,0,sizeof(vis));
        int num = bfs();
        if(num)printf("%d\n",num);
        else printf("NO\n");
    }
	return 0;
}