题目描述 Description
有两个无刻度标志的水壶，分别可装 x 升和 y 升 （ x,y 为整数且均不大于 100 ）的水。设另有一水 缸，可用来向水壶灌水或接从水壶中倒出的水， 两水壶间，水也可以相互倾倒。已知 x 升壶为空 壶， y 升壶为空壶。问如何通过倒水或灌水操作， 用最少步数能在x或y升的壶中量出 z （ z ≤ 100 ）升的水 来。
输入描述 Input Description
一行，三个数据，分别表示 x,y 和 z;
输出描述 Output Description
一行，输出最小步数 ,如果无法达到目标，则输出"impossible"
样例输入 Sample Input
3 22 1
样例输出 Sample Output
14
数据范围及提示 Data Size & Hint
广度优先搜索 深度优先搜索 迭代搜索 搜索
【DFS：】
#include<bits/stdc++.h>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const int N = 1e4+10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[][3]={ {0,0,1},{0,0,-1},{1,0,0},{-1,0,0},{0,1,0},{0,-1,0} };
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int f[200][200],a,b,z;
void dfs(int x,int y,int step)
{
if(f[x][y]!=0 && step+1>=f[x][y]) return ;
f[x][y]=step+1;
dfs(x,0,step+1);
dfs(0,y,step+1);
dfs(x,b,step+1);
dfs(a,y,step+1);
if(x+y<=a) dfs(x+y,0,step+1);
else dfs(a,x+y-a,step+1);
if(x+y<=b) dfs(0,x+y,step+1);
else dfs(x+y-b,b,step+1);
}
int main()
{
while(~scanf("%d%d%d",&a,&b,&z))
{
memset(f,0,sizeof(f));
int ans=INF;
dfs(0,0,0);
for(int i=0;i<=a;i++)
{
if(f[i][z]!=0)
{
if(f[i][z]<ans)
ans=f[i][z];
}
}
for(int i=0;i<=b;i++)
{
if(f[z][i]!=0)
{
if(f[z][i]<ans)
ans=f[z][i];
}
}
if(ans==INF) printf("impossible
");
else printf("%d
",ans-1);
}
}
【BFS】：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int x,y,target;
struct state{
int x,y,step;
}f;
queue<state> q;
bool vis[233][233];
int bfs()
{
q.push(f);
vis[f.x][f.y]=1;
while(q.size())
{
f=q.front(); q.pop();
if(f.x==target||f.y==target) return f.step;
if(f.x<x&&vis[x][f.y]==0) //x倒满
{
q.push((state){x,f.y,f.step+1});
vis[x][f.y]=1;
}
if(f.x&&vis[0][f.y]==0) //x倒空
{
q.push((state){0,f.y,f.step+1});
vis[0][f.y]=1;
}
if(f.y<y&&vis[f.x][y]==0) //y倒满
{
q.push((state){f.x,y,f.step+1});
vis[f.x][y]=1;
}
if(f.y&&vis[f.x][0]==0) //y倒空
{
q.push((state){f.x,0,f.step+1});
vis[f.x][0]=1;
}
if(f.x>=y-f.y&&vis[f.x-(y-f.y)][y]==0)//x->y
{
q.push((state){f.x-(y-f.y),y,f.step+1});
vis[f.x-(y-f.y)][y]=1;
}
if(f.x<y-f.y&&vis[0][f.x+f.y]==0)//x->y
{
q.push((state){0,f.x+f.y,f.step+1});
vis[0][f.x+f.y]=1;
}
if(f.y>=x-f.x&&vis[x][f.y-(x-f.x)]==0)//y->x
{
q.push((state){x,f.y-(x-f.x),f.step+1});
vis[x][f.y-(x-f.x)]=1;
}
if(f.y<x-f.x&&vis[f.x+f.y][0]==0)//y->x
{
q.push((state){f.x+f.y,0,f.step+1});
vis[f.x+f.y][0]=1;
}
}
return -1;
}
int main()
{
scanf("%d%d%d",&x,&y,&target);
int ans=bfs();
if(ans!=-1) printf("%d",ans);
else printf("impossible");
return 0;
}