【ACM】杭电 hdu 1482 Counterfeit Dollar

/* THE PROGRAM IS MADE BY PYY *//*----------------------------------------------------------------------------//    Copyright (c) 2011 panyanyany All rights reserved.    URL   : http://acm.hdu.edu.cn/showproblem.php?pid=1482    Name  : 1482 Counterfeit Dollar    Date  : Sunday, January 22, 2012    Time Stage : 2 hours    Result: 52857192012-01-22 11:47:15Accepted14820MS236K2462 BC++pyyTest Data :Review :不是什么难题，慢慢推敲一下就出来了。首先，基本原理是，把全部倾斜的天平摆成一个方向，设为左低右高，那么假币肯定在任意一边。设它在左边，则左边3组中必然都存在假币，所以只要枚举左边所有倾斜组，得到共有的那个硬币，便是假币。如果左边找不到所有组共有的硬币，则在右边找。如果遇到平衡的天平，那么此天平两边的硬币都是真币，就可以在枚举的时候过滤掉这些真币了，同时也要减少枚举的组数。比如示例中，有两组真币，那么只枚举一组就行了。//----------------------------------------------------------------------------*/#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <string>using namespace std ;#define INF        (-1)#define MAXN       ('L'-'A'+2)int evenCnt ;int getsum (bool side[MAXN][3], bool even[MAXN]){int i, j, sum ;for (i = 0 ; i < 12 ; ++i){sum = 0 ;if (even[i])// 不枚举真币continue ;for (j = 0 ; j < 3 ; ++j)sum += side[i][j] ;if (sum == 3 - evenCnt)// 只枚举倾斜的天平break ;}return i ;}int main (){int i, j ;int tcase, k, id ;string a, b, c, lf, rh ;char *pWei ;bool even[MAXN], left[MAXN][3], right[MAXN][3] ;while (scanf ("%d", &tcase) != EOF){k = 0 ;while (k++ < tcase){MEM (even, 0) ;MEM (left, 0) ;MEM (right, 0) ;evenCnt = 0 ;for (i = 0 ; i < 3 ; ++i){cin >> a >> b >> c ;if (c[0] == 'e')// 记录真币{++evenCnt ;for (j = 0 ; j < a.size() ; ++j){even[a[j]-'A'] = true ;}for (j = 0 ; j < b.size() ; ++j){even[b[j]-'A'] = true ;}continue ;}else if (c[0] == 'u')lf = a, rh = b ;elself = b, rh = a ;for (j = 0 ; j < lf.size() ; ++j)left[lf[j]-'A'][i] = true ;for (j = 0 ; j < rh.size() ; ++j)right[rh[j]-'A'][i] = true ;}id = getsum (left, even) ;pWei = "heavy" ;if (id == 12){id = getsum (right, even) ;pWei = "light" ;}printf ("%c is the counterfeit coin and it is %s.\n", id+'A', pWei) ;}}return 0 ;}