Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input
3 5 4
Sample Output
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)思路BFS,用两个罐子的剩余水量作为状态标记
代码：
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
int A, B, C;
int vis[105][105];
struct V{
int a;
int b;
int step;
string ans;
}p,pp;
queue <V> s;
bool bfs()
{
while (!s.empty())
s.pop();
vis[0][0] = 1;
p.a = 0;
p.b = 0;
p.step = 0;
s.push(p);
while (!s.empty())
{
p = s.front();
s.pop();
if (p.a == C || p.b == C)
return true;
int aa, bb;
for (int i = 1; i <= 6; i++){
if (1 == i)
aa = A, bb = p.b;
else if (2 == i)
aa = p.a, bb = B;
else if (3 == i){
if (p.a + p.b > A)
bb = p.a + p.b - A, aa = A;
else
bb = 0, aa = p.a + p.b;
}
else if (4 == i){
if (p.a + p.b > B)
aa = p.a + p.b - B, bb = B;
else
aa = 0, bb = p.a + p.b;
}
else if (5 == i)
aa = 0, bb = p.b;
else
aa = p.a, bb = 0;
if (!vis[aa][bb]){
vis[aa][bb] = 1;
pp.a = aa;
pp.b = bb;
pp.step = p.step + 1;
char c = i + '0';
pp.ans = "";
pp.ans += (p.ans + c);
s.push(pp);
}
}
}
return false;
}
void output(char c)
{
if ('1' == c)
puts("FILL(1)");
else if ('2' == c)
puts("FILL(2)");
else if ('3' == c)
puts("POUR(2,1)");
else if ('4' == c)
puts("POUR(1,2)");
else if ('5' == c)
puts("DROP(1)");
else
puts("DROP(2)");
}
int main()
{
while (scanf("%d%d%d", &A, &B, &C) != EOF)
{
memset(vis, 0, sizeof(vis));
if (bfs()){
printf("%d\n", p.step);
for (int i = 0; i < p.step; i++)
output(p.ans[i]);
}
else
puts("impossible");
}
return 0;
}
原文地址：http://blog.csdn.net/u012964281/article/details/38143561
