原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1495
一:分析
思路:预处理m < n < s,以后处理方便点
初始状态,m,n杯中可乐体积为0,s杯中体积为s;
然后分六种情况:
1, s 倒 m
2, s 倒 n
3, m 倒 n
4, m 倒 s
5, n 倒 m
6, n 倒 s
直到n,s杯中的可乐能等分(此时m杯中体积为0)为止,若不能等分,则输出 NO
二:AC代码
#define _CRT_SECURE_NO_DEPRECATE
#define _CRT_SECURE_CPP_OVERLOAD_STANDARD_NAMES 1
#include
#include
#include
using namespace std;
#define maxn 101
bool visited[maxn][maxn];
int m, n, s, si, sj;
struct node
{
int x, y, all, t; //x,y,all分别表示m,n,s杯中可乐的体积,t表示倒了多少次
};
void BFS()
{
queue que;
memset(visited, false, sizeof(visited));
node p, q;
p.x = 0, p.y = 0, p.t = 0, p.all = s;
que.push(p);
visited[p.x][p.y] = true;
while (!que.empty())
{
p = que.front();
que.pop();
if (p.y == p.all && p.y == s / 2)
{
printf("%d\n", p.t);
return;
}
if (p.all + p.x > m) //s倒m
{
q.x = m, q.y = p.y, q.all = p.all + p.x - m, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
else
{
q.x = p.all + p.x, q.y = p.y, q.all = 0, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
if (p.all + p.y > n) //s倒n
{
q.x = p.x, q.y = n, q.all = p.all + p.y - n, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
else
{
q.x = p.x, q.y = p.all + p.y, q.all = 0, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
if (p.x + p.y > n) //m倒n
{
q.x = p.x + p.y - n, q.y = n, q.all = p.all, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
else
{
q.x = 0, q.y = p.x + p.y, q.all = p.all, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
q.all = p.all + p.x, q.x = 0, q.y = p.y, q.t = p.t + 1; //m倒s
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
if (p.x + p.y > m)
{
q.y = p.y + p.x - m, q.x = m, q.all = p.all, q.t = p.t + 1;//n倒m
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
else
{
q.x = p.x + p.y, q.y = 0, q.all = p.all, q.t = p.t + 1;
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
q.all = p.all + p.y, q.x = p.x, q.y = 0, q.t = p.t + 1; //n倒s
if (!visited[q.x][q.y])
que.push(q), visited[q.x][q.y] = true;
}
printf("NO\n");
}
int main()
{
while (scanf("%d%d%d", &s, &m, &n) && (s || m || n))
{
if (s % 2)
{
printf("NO\n");
continue;
}
if (m > n) swap(m, n);
BFS();
}
return 0;
}