POJ 2408 Anagram Groups 排序到极致

Anagram Groups

Description

World-renowned Prof. A. N. Agram‘s current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text, you are to find the largest anagram groups.

A text is a sequence of words. A word w is an anagram of a word v if and only if there is some permutation p of character positions that takes w to v. Then, w and v are in the same anagram group. The size of an anagram group is the number of words in that group.
Find the 5 largest anagram groups.

Input

The input contains words composed of lowercase alphabetic characters, separated by whitespace(or new line). It is terminated by EOF. You can assume there will be no more than 30000 words.

Output

Output the 5 largest anagram groups. If there are less than 5 groups, output them all. Sort the groups by decreasing size. Break ties lexicographically by the lexicographical smallest element. For each group output, print its size and its member words. Sort
the member words lexicographically and print equal words only once.

Sample Input

undisplayed
trace
tea
singleton
eta
eat
displayed
crate
cater
carte
caret
beta
beat
bate
ate
abet

Sample Output

Group of size 5: caret carte cater crate trace .
Group of size 4: abet bate beat beta .
Group of size 4: ate eat eta tea .
Group of size 1: displayed .
Group of size 1: singleton .

Source

Ulm Local 2000

对strcmp();认识不到位WA一天了 哎

AC:代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
struct my{
    char w1[30];
    char w2[30];
    int size;
    int len;
    void put(){
    printf("w1: %s  w2: %s  len: %d  size:  %d\n",w1,w2,len,size);
    }
};
struct on{
    int isize;
    int start;
    int end;
    void put(){
    printf("start: %d end: %d isize: %d\n",start,end,isize);
    }
};
on z[30010];
my me[30010];
bool cmp1(my a,my b){
    if(a.len==b.len){
        if(strcmp(a.w2,b.w2)==0)
            return strcmp(a.w1,b.w1)<0?true:false;
        else return strcmp(a.w2,b.w2)<0?true:false;
    }
    return a.len>b.len;
}
bool cmp2(on a,on b){
    if(a.isize==b.isize){
        return strcmp(me[a.start].w1,me[b.start].w1)<0?true:false;
    }
    return a.isize>b.isize;
}
int main(){
    char temp[30];
    int n=0;
    while(scanf("%s",temp)==1){
        int l=strlen(temp);
        strcpy(me[n].w1,temp);
        me[n].len=l;
        me[n].size=1;
        sort(temp,temp+l);
        strcpy(me[n].w2,temp);
        n++;
    }
    ///for(int i=0;i<n;++i)me[i].put();
    sort(me,me+n,cmp1);
    ///cout<<"\n\n";
   /// for(int i=0;i<n;++i)me[i].put();
    int nz=0;
    for(int i=0,add,t;i<n;++i){
        add=0;t=i;
        while(!strcmp(me[i].w2,me[i+1].w2)&&i<n){
            add++;i++;
        }
        for(int j=t;j<=i;j++)me[j].size+=add;
        z[nz].start=t;
        z[nz].end=i;
        z[nz].isize=add+1;
        nz++;
    }
    sort(z,z+nz,cmp2);
    ///cout<<"\n\n";
    ///for(int i=0;i<n;++i)me[i].put();
    ///cout<<"\n\n";
   /// for(int i=0;i<nz;++i)z[i].put();
    int loop=5;
    int i=0;
    while(loop--&&i<n&&i<nz){
        printf("Group of size %d: ",z[i].isize);
        printf("%s ",me[z[i].start].w1);
        for(int j=z[i].start+1;j<=z[i].end;++j){
            if(strcmp(me[j].w1,me[j-1].w1)!=0)
                printf("%s ",me[j].w1);
        }
        printf(".\n");
        i++;
    }
    return 0;
}

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