标签：des blog io os 使用 ar for sp div
Time Limit: 1 secs, Memory Limit: 32 MB
It‘s preseason and the local newspaper wants to publish a preseason ranking of the teams in the local amateur basketball league. The teams are the Ants, the Buckets, the Cats, the Dribblers, and the Elephants. When Scoop McGee, sports editor of the paper, gets the rankings from the selected local experts down at the hardware store, he‘s dismayed to find that there doesn‘t appear to be total agreement and so he‘s wondering what ranking to publish that would most accurately reflect the rankings he got from the experts. He’s found that finding the median ranking from among all possible rankings is one way to go.
The median ranking is computed as follows: Given any two rankings, for instance ACDBE and ABCDE, the distance between the two rankings is defined as the total number of pairs of teams that are given different relative orderings. In our example, the pair B, C is given a different ordering by the two rankings. (The first ranking has C above B while the second ranking has the opposite.) The only other pair that the two rankings disagree on is B, D; thus, the distance between these two rankings is 2. The median ranking of a set of rankings is that ranking whose sum of distances to all the given rankings is minimal. (Note we could have more than one median ranking.) The median ranking may or may not be one of the given rankings.
Suppose there are 4 voters that have given the rankings: ABDCE, BACDE, ABCED and ACBDE. Consider two candidate median rankings ABCDE and CDEAB. The sum of distances from the ranking ABCDE to the four voted rankings is 1 + 1 + 1 + 1 = 4. We‘ll call this sum the value of the ranking ABCDE. The value of the ranking CDEAB is 7 + 7 + 7 + 5 = 26.
It turns out that ABCDE is in fact the median ranking with a value of 4.
There will be multiple input sets. Input for each set is a positive integer n on a line by itself, followed by n lines (n no more than 100), each containing a permutation of the letters A, B, C, D and E, left-justified with no spaces. The final input set is followed by a line containing a 0, indicating end of input.
Output for each input set should be one line of the form:
ranking is the median ranking with value value.
Of course ranking should be replaced by the correct ranking and value with the correct value. If there is more than one median ranking, you should output the one which comes first alphabetically.
4 ABDCE BACDE ABCED ACBDE 0
ABCDE is the median ranking with value 4.
#include <iostream>
#include <map>
#include <vector>
using namespace std;
//计算对应单个rankings的一个medianrank的distance
int rankdistance(string rank, map<char, int> order) {
int sub_value = 0;
for (int i = 0; i < rank.size(); i++) {
for (int j = i+1; j < rank.size(); j++) {
if (order[rank[i]] > order[rank[j]])//队排名越前，order[rank[i]]的值越小，此处为若单个rankings中的某个队排名比medianrank的高，则产生distanc
sub_value++;
}
}
return sub_value;
}
//计算对应一个rankings数组的一个medianrank的value
int rankvalue(vector<string> rankings, string rank) {
int value = 0;
for (int i = 0; i < rankings.size(); i++) {//对每个rankings数组的rank对medianrank计算其sub_value，累加成对所以rankings的value
map<char, int> order; //此步很关键，使用map来记录rankings数组中的每个rank中的每个队的排名，以便计算distance
for (int j = 0; j < rank.size(); j++) {
order[rankings[i].at(j)] = j;
}
value += rankdistance(rank, order);
}
return value;
}
//枚举所有可能的medianrank并计算其对应一个rankings数组的value，同时选择value值最小且字母顺序最小的medianrank
void perminate(vector<string> rankings, string source, int &minvalue, string &result, int index) {//记得minvalue和result要设为引用类型，才能使43,44行起作用
if (index == source.length()-1) {
string temp;
for (int i = 0; i < source.length(); i++) {
temp += source[i];
}
int tempvalue = rankvalue(rankings, temp);
//对所有可能的medinarank的value进行比较，选择value最小的medianrank ,此处使用<而非<=保证了相同的value的medianrank选择字母顺序最小的
if (tempvalue < minvalue) {
minvalue = tempvalue;
result = temp;
}
} else {
//由于程序执行时输入为ABCDE作为source，所以以下实现的该ABCDE五个字母的全排列为升序的
for (int i = index; i < source.length(); i++) {
swap(source[index], source[i]);
perminate(rankings, source, minvalue, result, index+1);
swap(source[index], source[i]);
}
}
}
int main() {
int cases;
while (cin >> cases && cases) {
string str;
vector<string> rankings;
for (int i = 0; i < cases; i++) {
cin >> str;
rankings.push_back(str);
}
string source = "ABCDE";
int minvalue = 1000;//由于5个字母的最大distance为10，所以minvalue的初始值设为很大
string result;
perminate(rankings, source, minvalue, result, 0);
cout << result << " is the median ranking with value " << minvalue <<"." <<endl;
}
return 0;
}
sicily 1006 team rankings 枚举解题
标签：des blog io os 使用 ar for sp div
原文地址：http://www.cnblogs.com/xieyizun-sysu-programmer/p/4005284.html
