题目链接：poj 2408 Anagram Groups

题目大意：给定若干个字符串，将其分组，按照组成元素相同为一组，输出数量最多的前5组，每组按照字典序输出所

有字符串。数量相同的输出字典序较小的一组。

解题思路：将所有的字符串统计字符后hash，排序之后确定每组的个数并且确定一组中字典序最小的字符串。根据个数

以及字符串对组进行排序。

#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

const int maxn = 30005;
const int maxm = 30;
const int X = 30;
typedef unsigned long long ll;
typedef pair<ll,int> pii;

int N, M, E;
vector<pii> vec, grop;
vector<int> g[maxn];
char word[maxn][maxm], st[maxn][maxm];

inline ll Hash(char* s) {
    int len = strlen(s), c[maxm];
    memset(c, 0, sizeof(c));

    for (int i = 0; i < len; i++)
        c[s[i]-‘a‘]++;

    ll ret = 0;
    for (int i = 0; i < 26; i++)
        ret = ret * X + c[i];
    return ret;
}

inline bool cmp (const pii& a, const pii& b) {
    if (a.second == b.second)
        return strcmp(st[a.first], st[b.first]) < 0;
    return a.second > b.second;
}

inline bool sort_by(const int& a, const int& b) {
    return strcmp(word[a], word[b]) < 0;
}

int main () {
    N = M = E = 0;
    vec.clear();
    grop.clear();

    while (scanf("%s", word[N]) == 1) {
        ll key = Hash(word[N]);
        vec.push_back(make_pair(key, N));
        N++;
    }
    sort(vec.begin(), vec.end());

    int cnt = 0;
    ll pre = -1;

    for (int i = 0; i < vec.size(); i++) {
        int idx = vec[i].second;
        if (vec[i].first != pre) {
            if (cnt)
                grop.push_back(make_pair(M++, cnt));
            cnt = 0;
            g[M].clear();
            pre = vec[i].first;
            strcpy(st[M], word[idx]);
        }

        cnt++;
        g[M].push_back(idx);
        if (strcmp(word[idx], st[M]) < 0)
            strcpy(st[M], word[idx]);
    }

    if (cnt)
        grop.push_back(make_pair(M++, cnt));
    sort(grop.begin(), grop.end(), cmp);

    for (int i = 0; i < min(5, (int)grop.size()); i++) {
        printf("Group of size %d: ", grop[i].second);
        int x = grop[i].first;
        sort(g[x].begin(), g[x].end(), sort_by);
        for (int j = 0; j < g[x].size(); j++) {
            if (j == 0 || strcmp(word[g[x][j-1]], word[g[x][j]]))
                printf("%s ", word[g[x][j]]);
        }
        printf(".\n");
    }
    return 0;
}

poj 2408 Anagram Groups(hash)