[SOJ1006] Team Rankings

用枚举法实现
代码如下：
//我是先用全排列生成了全部的串，再写的
全排列生成代码，点击可看
~~枚举的时候用了减枝，所以可能会稍微快点~~
但是emmm
~~看完师兄们的解法只需要0.00sec之后，我这个0.02sec的菜鸡就不知道该说什么好了~~

代码如下:（会比较好理解）

#include <iostream>
#include <vector>
using namespace std;
#include <cstring>
#include <algorithm>
string strSet[] = {
"ABCDE",
"ABCED",
"ABDCE",
"ABDEC",
"ABECD",
"ABEDC",
"ACBDE",
"ACBED",
"ACDBE",
"ACDEB",
"ACEBD",
"ACEDB",
"ADBCE",
"ADBEC",
"ADCBE",
"ADCEB",
"ADEBC",
"ADECB",
"AEBCD",
"AEBDC",
"AECBD",
"AECDB",
"AEDBC",
"AEDCB",
"BACDE",
"BACED",
"BADCE",
"BADEC",
"BAECD",
"BAEDC",
"BCADE",
"BCAED",
"BCDAE",
"BCDEA",
"BCEAD",
"BCEDA",
"BDACE",
"BDAEC",
"BDCAE",
"BDCEA",
"BDEAC",
"BDECA",
"BEACD",
"BEADC",
"BECAD",
"BECDA",
"BEDAC",
"BEDCA",
"CABDE",
"CABED",
"CADBE",
"CADEB",
"CAEBD",
"CAEDB",
"CBADE",
"CBAED",
"CBDAE",
"CBDEA",
"CBEAD",
"CBEDA",
"CDABE",
"CDAEB",
"CDBAE",
"CDBEA",
"CDEAB",
"CDEBA",
"CEABD",
"CEADB",
"CEBAD",
"CEBDA",
"CEDAB",
"CEDBA",
"DABCE",
"DABEC",
"DACBE",
"DACEB",
"DAEBC",
"DAECB",
"DBACE",
"DBAEC",
"DBCAE",
"DBCEA",
"DBEAC",
"DBECA",
"DCABE",
"DCAEB",
"DCBAE",
"DCBEA",
"DCEAB",
"DCEBA",
"DEABC",
"DEACB",
"DEBAC",
"DEBCA",
"DECAB",
"DECBA",
"EABCD",
"EABDC",
"EACBD",
"EACDB",
"EADBC",
"EADCB",
"EBACD",
"EBADC",
"EBCAD",
"EBCDA",
"EBDAC",
"EBDCA",
"ECABD",
"ECADB",
"ECBAD",
"ECBDA",
"ECDAB",
"ECDBA",
"EDABC",
"EDACB",
"EDBAC",
"EDBCA",
"EDCAB",
"EDCBA"
};struct Node{char x,y;Node(char a='A',char b='A'):x(a),y(b){};
}; int label[120];//总共有120种可能排法，只需要找到最简单的那个就好了
int minAns = 1<<30;
string ans;
string DataSet[101];
int n; int jiao(vector<Node> v1,vector<Node> v2){int a = 0;for (int i = 0; i < v1.size(); ++i) {for (int j = 0; j < v2.size(); ++j){if (v1[i].x == v2[j].x && v1[i].y == v2[j].y){a++;break;}}}return a;
}
int cHelp(int p,int q){//返回第p个序列和第q个序列之间的差距// p在strSet中找，q在DataSet中找 vector<Node> v1,v2;//v1放的是strSet[p]的二元子串；v2放的是v2放的是DataSet[q]的二元子串 for (int i = 0; i < 4;++i){for (int j = i + 1; j < 5; ++j){v1.push_back(Node(strSet[p][i], strSet[p][j]));v2.push_back(Node(DataSet[q][i], DataSet[q][j]));}}// sort and  int num = jiao(v1,v2);return 10 - num;
}
// calculate
void calculate(int temp){ // temp在strSet中  //计算第temp个字符的sum和//实现一个减枝，如果目前加的和已经大于了当前最小的那个，就可以不用来看了for (int j = 0; j < n; ++j){label[temp] +=  cHelp(temp, j);if (label[temp] > minAns) {break;} // 减枝部分 }if (label[temp] < minAns){ans = strSet[temp];minAns = label[temp]; // 这是数值更新部分 }
}
int main(){while (cin >> n && n){for (int i = 0; i < n; ++i){cin >> DataSet[i];}memset(label,0,sizeof(label));//标记复原为0; minAns = 1<<30;ans = "";//初始化 for (int i = 0; i < 120; ++i){calculate(i);} cout << ans<<" is the median ranking with value "<<minAns<<"."<< endl;}
}

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