ZOJ 1005 Jugs

题意：两个罐子倒水

题解：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int a, b, n;
bool vis[1010][1010], flag;
int res[10100], step;

void print()
{
    for(int i = 0; i < step; i++)
    {
        switch(res[i])
        {
        case 1: printf("fill A\n"); break;
        case 2: printf("fill B\n"); break;
        case 3: printf("empty A\n"); break;
        case 4: printf("empty B\n"); break;
        case 5: printf("pour A B\n"); break;
        case 6: printf("pour B A\n"); break;
        case 7: printf("success\n"); break;
        }
    }
}

void dfs(int x, int y)
{
    if(flag) return;
    if(x == n || y == n)
    {
        flag = true;
        res[step++] = 7;
        print();
        return;
    }

    if(a > x && !vis[a][y] && !flag)
    {
        res[step++] = 1;
        vis[a][y] = true;
        dfs(a, y);
        step--;
    }

    if(b > y && !vis[x][b] && !flag)
    {
        res[step++] = 2;
        vis[x][b] = true;
        dfs(x, b);
        step--;
    }

    if(x != 0 && !vis[0][y] && !flag)
    {
        res[step++] = 3;
        vis[0][y] = true;
        dfs(0, y);
        step--;
    }

    if(y != 0 && !vis[x][0] && !flag)
    {
        res[step++] = 4;
        vis[x][0] = true;
        dfs(x, 0);
        step--;
    }

    int tmp = min(x, b - y);
    if(!vis[x-tmp][y+tmp] && !flag)
    {
        res[step++] = 5;
        vis[x-tmp][y+tmp] = true;
        dfs(x - tmp, y + tmp);
        step--;
    }

    tmp = min(a - x, y);
    if(!vis[x+tmp][y-tmp] && !flag)
    {
        res[step++] = 6;
        vis[x+tmp][y-tmp] = true;
        dfs(x + tmp, y - tmp);
        step--;
    }
}

int main()
{
    while(scanf("%d%d%d",&a,&b,&n) != EOF)
    {
        flag = false;
        memset(vis, 0, sizeof(vis));
        step = 0;
        vis[0][0] = true;
        dfs(0, 0);
    }
    return 0;
}