uva 10132 File Fragmentation(检索）

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题目连接：10132 File Fragmentation

题目大意：有n个有0， 1 组成的文件，每个文件被随意分两片，现在给出所有的2n碎片，要求找出原有的序列。

解题思路：右两端寻找，最短的一定匹配最长的。

#include <stdio.h>#include <string.h>#include <string>#include <algorithm>#include <iostream>using namespace std;const int N = 300;char str[N];string file[N];int vis[N], len, n;string order;bool cmp(const string &a, const string &b) {    return a.length() < b.length();}bool solved(int cur) {    if (cur >= n / 2)return true;    vis[cur] = 1;    int l = file[cur].length();    for (int i = n - 1; i >= 0; i--) {if (vis[i])   continue;if (l + file[i].length() > len)   return true;if (l + file[i].length() < len)   return false;string p, q;p = file[cur] + file[i];q = file[i] + file[cur];if (p == order || q == order) {    vis[i] = 1;    if (solved(cur + 1))return true;    vis[i] = 0;}    }    vis[cur] = 0;    return false;}int main() {    int cas, sum, max;    scanf("%d%*c%*c", &cas);    while (cas--) {sum = n = 0;memset(vis, 0, sizeof(vis));while (gets(str)) {    if (!str[0])break;    file[n] = str;    sum += file[n++].length();}sort(file, file + n, cmp);len = sum * 2 / n;vis[0] = 1;for (int i = n - 1; i >= 0; i--) {    vis[i] = 1;    order.clear();    order = file[0] + file[i];    if (solved(1))  break;    order.clear();    order = file[i] + file[0];    if (solved(1))  break;    vis[i] = 0;}cout << order << endl;if (cas)    cout << endl;    }    return 0;}