HDU 1495 非常可乐 (BFS)

题意：略。

析：由于只有三只杯子，那么我们可以用两个杯子的状态，那么第三只的状态也可以确定下来，每次倒水，要么全倒过去，要么把那个杯子倒满。

注意题意是要保证最后是两个一样多。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int d[maxn][maxn];
int all;

struct Node{
  int a, b;
  Node(int aa, int bb) : a(aa), b(bb) { }
};

bool judge(Node &u){
  if(u.a * 2 == all)  return u.b == 0 || (all - u.a - u.b) == 0;
  if(u.b * 2 == all)  return u.a == 0 || (all - u.a - u.b) == 0;
  if((all - u.a - u.b) * 2 == all)  return u.a == 0 || u.b == 0;
  return false;
}

void update(int &x, int &y, Node &u, queue<Node> &q){
  d[x][y] = d[u.a][u.b] + 1;
  q.push(Node(x, y));
}

int bfs(){
  queue<Node> q;
  q.push(Node(all, 0));
  memset(d, -1, sizeof d);
  d[all][0] = 0;

  while(!q.empty()){
    Node u = q.front();  q.pop();
    if(judge(u))  return d[u.a][u.b];
    // pour the first to the second or not
    int t = min(u.a, n - u.b);
    int x = u.a - t, y = u.b + t;
    if(d[x][y] == -1)  update(x, y, u, q);
    t = min(all-u.a, u.b);
    x = u.a + t, y = u.b - t;
    if(d[x][y] == -1)  update(x, y, u, q);
    // pour the first to the third or not
    t = min(u.a, m - (all-u.a-u.b));
    x = u.a - t, y = u.b;
    if(d[x][y] == -1)  update(x, y, u, q);
    t = min(all-u.a, all-u.a-u.b);
    x = u.a + t;
    if(d[x][y] == -1)  update(x, y, u, q);
    //pour the second to the third or not
    t = min(u.b, m - (all-u.a-u.b));
    x = u.a, y = u.b - t;
    if(d[x][y] == -1)  update(x, y, u, q);
    t = min(n-u.b, all-u.a-u.b);
    y = u.b + t;
    if(d[x][y] == -1)  update(x, y, u, q);
  }
  return -1;
}

int main(){
  while(scanf("%d %d %d", &all, &n, &m) == 3 && all + n + m){
    all <<= 1, n <<= 1, m <<= 1;
    int ans = bfs();
    if(ans == -1)  printf("NO
");
    else printf("%d
", ans);
  }
  return 0;
}