hdu 1485 非常可乐 题解

参见小白书130页-131页，关键是在状态的存储，一开始用L*100+M*10+S存储每一个状态,L,M,S分别表示第一个第二个第三个杯子所含的可乐的量，不能过，因为这样的编码有重复。

#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;}