POJ 3414 Pots

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题意：有两个水壶，体积为A、B和无限量的水，在这两个水壶里相互倒水，判断怎么倒能到C体积的水

链接：http://poj.org/problem?id=3414

思路：dfs，每个水壶作为节点建图进行搜索

注意点：无

以下为AC代码：

Run IDUserProblemResultMemoryTimeLanguageCode LengthSubmit Time13918617luminous113414Accepted780K0MSG++3598B2015-02-28 15:18:54

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;//const double pi = acos(-1);//const double eps = 1e-10;//const int dir[[4][2] = { 1,0, -1,0, 0,1, 0,-1 };int b, c, a;bool vis[105][105];struct node{    int b, c;    string str;    node(){}    node ( int _b, int _c, string _str ) : b(_b), c(_c), str(_str) {}};string ans;inline bool judge ( const node &tmp ){    if ( tmp.b == a || tmp.c == a ){        ans = tmp.str;        return true;    }    else{        return false;    }}bool solve(){    clr ( vis, 0 );    queue<node> q;    q.push ( node( 0, 0, "" ) );    while ( ! q.empty() ){        node tmp = q.front();        q. pop();        if ( judge ( tmp ) ){            return true;        }        // fill 1 '1        if ( tmp.b != b ){            if ( vis[b][tmp.c] == 0 ){                vis[b][tmp.c] = 1;                q.push ( node ( b, tmp.c, tmp.str + "1" ) );            }        }        // fill 2 '2        if ( tmp.c != c ){            if ( vis[tmp.b][c] == 0 ){                vis[tmp.b][c] = 1;                q.push ( node ( tmp.b, c, tmp.str + "2" ) );            }        }        // drop 1 '3        if ( tmp.b ){            if ( vis[0][tmp.c] == 0 ){                vis[0][tmp.c] = 1;                q.push ( node ( 0, tmp.c, tmp.str + "3" ) );            }        }        // drop 2 '4        if ( tmp.c ){            if ( vis[tmp.b][0] == 0 ){                vis[tmp.b][0] = 1;                q.push ( node ( tmp.b, 0, tmp.str + "4" ) );            }        }        // pour12 '5        if ( tmp.b && tmp.c != c ){            int t = min ( tmp.b, c - tmp.c );            if ( vis[tmp.b-t][tmp.c+t] == 0 ){                vis[tmp.b-t][tmp.c+t] = 1;                q.push ( node ( tmp.b-t, tmp.c+t, tmp.str + "5" ) );            }        }        // pour21 '6        if ( tmp.c && tmp.b != b ){            int t = min ( tmp.c, b - tmp.b );            if ( vis[tmp.b+t][tmp.c-t] == 0 ){                vis[tmp.b+t][tmp.c-t] = 1;                q. push ( node ( tmp.b+t, tmp.c-t, tmp.str + "6" ) );            }        }    }    return 0;}void print (){    cout << ans.size() << endl;    for ( int i = 0; i < ans.size(); i ++ ){        if ( ans[i] == '1' )            cout << "FILL(1)" << endl;        if ( ans[i] == '2' )            cout << "FILL(2)" << endl;        if ( ans[i] == '3' )            cout << "DROP(1)" << endl;        if ( ans[i] == '4' )            cout << "DROP(2)" << endl;        if ( ans[i] == '5' )            cout << "POUR(1,2)" << endl;        if ( ans[i] == '6' )            cout << "POUR(2,1)" << endl;    }}int main(){    ios::sync_with_stdio( false );    //while ( cin >> b >> c >> a ){        cin >> b >> c >> a;        if ( solve() ){            print();        }        else{            cout << "impossible" << endl;        }    //}    return 0;}

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