USACO training course Mother's Milk /// DFS（有点意思） oj10120

题目大意：

输入 A B C 为三个容器的容量

一开始A B是空的 C是满的

每一次倾倒只能在 盛的容器满 或 倒的容器空 时才停止

输出当A容器空时 C容器内剩余量的所有可能值

Sample Input

8 9 10
2 5 10

Sample Output

1 2 8 9 10
5 6 7 8 9 10

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
#include<math.h>
using namespace std;
int flag[25],vis[25][25],ans[25];
int a,b,c,len;
void DFS(int nowa,int nowc)
{
    if(vis[nowa][nowc]) return;
    vis[nowa][nowc]=1;

    if(nowa==0&&!flag[nowc])
        flag[nowc]=1, ans[len++]=nowc;

    int nowb=c-nowa-nowc;

    if(nowa&&nowc<c) ///a->c
        DFS(nowa-min(nowa,c-nowc),nowc+min(nowa,c-nowc));
    if(nowa&&nowb<b) ///a->b
        DFS(nowa-min(nowa,b-nowb),nowc);
    if(nowc&&nowa<a) ///c->a
        DFS(nowa+min(nowc,a-nowa),nowc-min(nowc,a-nowa));
    if(nowc&&nowb<b) ///c->b
        DFS(nowa,nowc-min(nowc,b-nowb));
    if(nowb&&nowa<a) ///b->a
        DFS(nowa+min(a-nowa,nowb),nowc);
    if(nowb&&nowc<c) ///b->c
        DFS(nowa,nowc+min(c-nowc,nowb));
}
int main()
{
    while(~scanf("%d%d%d",&a,&b,&c))
    {
        memset(ans,0,sizeof(ans));
        memset(vis,0,sizeof(vis));
        memset(flag,0,sizeof(flag));
        len=0;
        DFS(0,c);
        sort(ans,ans+len);
        for(int i=0;i<len-1;i++)
            printf("%d ",ans[i]);
        printf("%d\n",ans[len-1]);
    }

    return 0;
}