POJ 1789 卡车的历史(Truck History) 最小树

地址 http://poj.org/problem?id=1789

 

解析
aaaaaaa
baaaaaa
abaaaaa
aabaaaa

aaaaaaa 和 baaaaaa 差别度为1
abaaaaa 和 baaaaaa 差别度为2
其余同理

那么以字符串为点差别度作为连点的边权
就有下图(作画工具限制，双箭头看做无向图的一条线即可)

根据此无向图生成最小树，得到的最小权值和就是答案

#include <iostream>
#include <vector>
#include <memory.h>
#include <string>
#include <algorithm>


using namespace std;

/*
样例输入： 样例输出：
4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
7
baaaaaa
abaaaaa
aabaaaa
aaabaaa
aaaabaa
aaaaaba
aaaaaab
10
bbaaaaa
abbaaaa
aaabbaa
baaaaaa
abaaaaa
aabaaaa
aaabaaa
aaaabaa
aaaaaba
aaaaaab
10
bbbaaaa
abbbaaa
aaabbba
baaaaab
abaaaab
aabaaab
aaabaab
aaaabab
aaaaabb
aaaaaab
0

The highest possible quality is 1/3.
*/

int n;
const int N = 2010;
vector<string> vs;

int g[N][N];

const int INF = 0x3f3f3f3f;

int dist[N];
int st[N];

int prim()
{
    memset(dist, 0x3f, sizeof dist);
    int res = 0;
    for (int i = 0; i < n; i++) {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        if (i && dist[t] == INF) return INF;

        if (i) res += dist[t];
        st[t] = true;

        for (int j = 1; j <= n; j++) dist[j] = min(dist[j], g[t][j]);
    }

    return res;
}

int main()
{
    while (1) {
        cin >> n;
        if (n == 0) break;
        memset(g, 0x3f, sizeof(g));
        memset(st, 0, sizeof st);
        vs.clear();
        vs.push_back("no_use,place_holder");
        for (int i = 0; i < n; i++) {
            string s;
            cin >> s;
            for (int j = 1; j < vs.size(); j++) {
                int difCount = 0;
                for (int idx = 0; idx < 7; idx++) {
                    if (s[idx] != vs[j][idx]) difCount++;
                }
                g[j][vs.size()] = g[vs.size()][j] = difCount;
            }
            vs.push_back(s);
        }

    /*    for (int i = 1; i < vs.size(); i++) {
            for (int j = i + 1; j < vs.size(); j++) {
                int difCount = 0;
                for (int idx = 0; idx < 7; idx++) {
                    if (vs[i][idx] != vs[j][idx]) difCount++;
                }
                g[j][i] = g[i][j] = difCount;
            }
        }*/

        int ret = prim();

        cout << "The highest possible quality is 1/" << ret <<"."<< endl;
    }
}