jobdu1475 非常可乐
http://ac.jobdu.com/problem.php?pid=1457
- 题目描述：
-
大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。
- 输入：
-
三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。
- 输出：
-
如果能平分的话请输出最少要倒的次数，否则输出"NO"。
- 样例输入：
-
7 4 3 4 1 3 0 0 0
- 样例输出：
-
NO 3
#include <stdio.h>
#include <queue>
using namespace std;
struct N{
int a, b, c;
int t;
};
queue<N> Q;
bool mark[101][101][101];
void AtoB(int &a, int sa, int &b, int sb) {
if (sb - b >= a){
b += a;
a = 0;
}
else{
a -= sb - b;
b = sb;
}
}
int BFS(int s, int n, int m) {
while (Q.empty() == false) {
N now = Q.front();
Q.pop();
int a, b, c;
a = now.a;
b = now.b;
c = now.c;
AtoB(a, s, b, n);
if (mark[a][b][c] == false){
mark[a][b][c] = true;
N tmp;
tmp.a = a;
tmp.b = b;
tmp.c = c;
tmp.t = now.t + 1;
if (a == s / 2 && b == s / 2) return tmp.t;
if (c == s / 2 && b == s / 2) return tmp.t;
if (a == s / 2 && c == s / 2) return tmp.t;
Q.push(tmp);
}
a = now.a;
b = now.b;
c = now.c;
AtoB(b, n, a, s);
if (mark[a][b][c] == false){
mark[a][b][c] = true;
N tmp;
tmp.a = a;
tmp.b = b;
tmp.c = c;
tmp.t = now.t + 1;
if (a == s / 2 && b == s / 2) return tmp.t;
if (c == s / 2 && b == s / 2) return tmp.t;
if (a == s / 2 && c == s / 2) return tmp.t;
Q.push(tmp);
}
a = now.a;
b = now.b;
c = now.c;
AtoB(a, s, c, m);
if (mark[a][b][c] == false){
mark[a][b][c] = true;
N tmp;
tmp.a = a;
tmp.b = b;
tmp.c = c;
tmp.t = now.t + 1;
if (a == s / 2 && b == s / 2) return tmp.t;
if (c == s / 2 && b == s / 2) return tmp.t;
if (a == s / 2 && c == s / 2) return tmp.t;
Q.push(tmp);
}
a = now.a;
b = now.b;
c = now.c;
AtoB(c, m, a, s);
if (mark[a][b][c] == false){
mark[a][b][c] = true;
N tmp;
tmp.a = a;
tmp.b = b;
tmp.c = c;
tmp.t = now.t + 1;
if (a == s / 2 && b == s / 2) return tmp.t;
if (c == s / 2 && b == s / 2) return tmp.t;
if (a == s / 2 && c == s / 2) return tmp.t;
Q.push(tmp);
}
a = now.a;
b = now.b;
c = now.c;
AtoB(b, n, c, m);
if (mark[a][b][c] == false){
mark[a][b][c] = true;
N tmp;
tmp.a = a;
tmp.b = b;
tmp.c = c;
tmp.t = now.t + 1;
if (a == s / 2 && b == s / 2) return tmp.t;
if (c == s / 2 && b == s / 2) return tmp.t;
if (a == s / 2 && c == s / 2) return tmp.t;
Q.push(tmp);
}
a = now.a;
b = now.b;
c = now.c;
AtoB(c, m, b, n);
if (mark[a][b][c] == false){
mark[a][b][c] = true;
N tmp;
tmp.a = a;
tmp.b = b;
tmp.c = c;
tmp.t = now.t + 1;
if (a == s / 2 && b == s / 2) return tmp.t;
if (c == s / 2 && b == s / 2) return tmp.t;
if (a == s / 2 && c == s / 2) return tmp.t;
Q.push(tmp);
}
}
return -1;
}
int main(){
int s, n, m;
while (scanf("%d%d%d", &s, &n, &m) != EOF) {
if (s == 0) break;
if (s % 2 == 1){
puts("NO");
continue;
}
for (int i = 0; i <= s; i++){
for (int j = 0; j <= n; j++){
for (int k = 0; k <= m; k++){
mark[i][j][k] = false;
}
}
}
N tmp;
tmp.a = s;
tmp.b = 0;
tmp.c = 0;
tmp.t = 0;
while (Q.empty() == false) Q.pop();
Q.push(tmp);
mark[s][0][0] = true;
int rec = BFS(s, n, m);
if (rec == -1)
puts("NO");
else printf("%d\n", rec);
}
return 0;
}