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                                              File Fragmentation

The Problem

Your friend, a biochemistry major, tripped while carrying a tray of computer files through the lab. All of the files fell to the ground and broke. Your friend picked up all the file fragments and called you to ask for help putting them back together again.

Fortunately, all of the files on the tray were identical, all of them broke into exactly two fragments, and all of the file fragments were found. Unfortunately, the files didn‘t all break in the same place, and the fragments were completely mixed up by their fall to the floor.

You‘ve translated the original binary fragments into strings of ASCII 1‘s and 0‘s, and you‘re planning to write a program to determine the bit pattern the files contained.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Input will consist of a sequence of ``file fragments‘‘, one per line, terminated by the end-of-file marker. Each fragment consists of a string of ASCII 1‘s and 0‘s.

Output

Output is a single line of ASCII 1‘s and 0‘s giving the bit pattern of the original files. If there are 2N fragments in the input, it should be possible to concatenate these fragments together in pairs to make N copies of the output string. If there is no unique solution, any of the possible solutions may be output.

Your friend is certain that there were no more than 144 files on the tray, and that the files were all less than 256 bytes in size.

Sample Input

1

011
0111
01110
111
0111
10111

Sample Output

01110111

思路：暴力。把每个字符串与都链接一下，然后用map每个链接后的文件的数量，然后找出输入数量的一半以上的文件，则为原文件。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
using namespace std;
const int inf=0x3f3f3f3f;
char str[1010][310];
struct node {
    char site[310];
} x[10010];
int main()
{
    int T,i,j;
    char a[20],r1[1010],r2[1010];
    int cnt,t;
    scanf("%d",&T);
    getchar();
    gets(a);
    while(T--) {
        cnt=t=0;
        while(gets(str[cnt])!=NULL) {
            if(str[cnt][0]=='\0') break;
            cnt++;
        }
        map<string,int >q;
        for(i=0; i<cnt; i++)
            for(j=i+1; j<cnt; j++) {
                strcpy(r1,str[i]);
                strcat(r1,str[j]);
                if(!q[r1])
                    strcpy(x[t++].site,r1);
                q[r1]++;
                strcpy(r2,str[j]);
                strcat(r2,str[i]);
                if(!q[r2])
                    strcpy(x[t++].site,r2);
                q[r2]++;

            }
        for(i=0; i<t; i++)
            if(q[x[i].site]>=cnt/2) {
                printf("%s\n",x[i].site);
                break;
            }
            if(T)
                printf("\n");
    }
    return 0;
}

UVA 10132-File Fragmentation(map还原字符串)

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原文地址：http://blog.csdn.net/u013486414/article/details/43561067