【BFS】HDU 1495

直达–> HDU 1495 非常可乐

相似题联动–>POJ 3414 Pots

题意：中文题，不解释。

思路：三个杯子倒来倒去，最后能让其中两个平分即可。可能性六种。判定的时候注意第三个杯子不能有水，倒的时候也要注意别超过了倒进去的杯子的容积。

a->b || a->c || b->a || b->c || c->a || c->b

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int vis[105][105][105];
struct node{
    int s,n,m;
    int step;
};
bool check(int S,int N,int M){
    if(S==0&&(N==M))
        return true;
    if(N==0&&(S==M))
        return true;
    if(M==0&&(S==N))
        return true;
    return false;
}
int bfs(int S,int M,int N){
    queue<node>Q;
    node P,T;
    P.s = S;
    P.m = 0;
    P.n = 0;
    P.step = 0;
    vis[S][0][0] = 1;
    Q.push(P);
    while(Q.size()){
        P = Q.front();
        Q.pop();
        if(check(P.s,P.m,P.n)){
            return P.step;
        }
        if(P.s){
            if(P.s>N-P.n){
                T.s = P.s-(N-P.n);
                T.n = N;
                T.m = P.m;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            else{
                T.s = 0;
                T.n = P.s+P.n;
                T.m = P.m;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            if(P.s>M-P.m){
                T.s = P.s-(M-P.m);
                T.m = M;
                T.n = P.n;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            else{
                T.s = 0;
                T.m = P.s+P.m;
                T.n = P.n;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
        }
        if(P.m){
            if(P.m>N-P.n){
                T.m = P.m-(N-P.n);
                T.n = N;
                T.s = P.s;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            else{
                T.m = 0;
                T.n = P.n+P.m;
                T.s = P.s;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            if(P.m>S-P.s){
                T.m = P.m-(S-P.s);
                T.s = S;
                T.n = P.n;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            else{
                T.m = 0;
                T.s = P.s+P.m;
                T.n = P.n;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
        }
        if(P.n){
            if(P.n>S-P.s){
                T.n = P.n-(S-P.s);
                T.s = S;
                T.m = P.m;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            else{
                T.n = 0;
                T.s = P.s+P.n;
                T.m = P.m;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            if(P.n>M-P.m){
                T.n = P.n-(M-P.m);
                T.m = M;
                T.s = P.s;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
            else{
                T.n = 0;
                T.m = P.m+P.n;
                T.s = P.s;
                if(!vis[T.s][T.n][T.m]){
                    T.step=P.step+1;
                    Q.push(T);
                    vis[T.s][T.n][T.m] = 1;
                }
            }
        }
    }
    return -1;
}
int main(){
    int S,M,N;
    while(~scanf("%d%d%d",&S,&N,&M)){
        if(S==0&&N==0&&M==0){
            break;
        }
        if(S%2){
            printf("NO\n");
            continue;
        }
        memset(vis,0,sizeof(vis));
        int ans = bfs(S,M,N);
        if(ans==-1) printf("NO\n");
        else printf("%d\n",ans);
    }
    return 0;
}