Uva10132file fragmentation (Greedy)

Question: uva10132file fragmentation

N files are identical, but they are accidentally broken. The cracks in each file are different, and each file is damaged into two fragments. Each file fragment is represented by a binary number, and 2 * n fragments are given. Ask the files (n) that can be obtained by such fragments ). If the answer is not unique, you can give one of them.

Solution: Because every two fragments form a file and find the longest part, it must match the smallest file to form a file. Try every situation, and verify whether the remaining fragments can constitute the file.

Note that the input of this question is separated by spaces.

Code:

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 300;const int M = 150;int n, sum;char ans[N];int vis[M];struct Fragment {char str[N];int len;}f[M];int cmp (const Fragment & a, const Fragment &b) {return a.len > b.len;}bool judge (int a, int b) {memset(vis, 0, sizeof(vis));vis[a] = vis[b] = 1;char temp[N];for (int i = 0; i < n; i++) {if (vis[i])continue;vis[i] = 1;int j;for (j = i + 1; j < n; j++) {if (vis[j])continue;if (f[i].len + f[j].len == sum) {strcpy (temp, f[i].str);strcat (temp, f[j].str);if (strcmp (temp, ans) == 0) {vis[j] = 1;break;}strcpy (temp, f[j].str);strcat (temp, f[i].str);if (strcmp (temp, ans) == 0) {vis[j] = 1;break;}}}if (j == n)return 0;}return 1;}void solve () {int max = f[0].len;int min;for (int i = 0; i < n; i++)if (f[i].len == sum - max) {min = i;break;}for (int i = 0; i < n; i++) {if (f[i].len != max)return;for (int j = min; j < n; j++) {if (f[j].len != f[min].len)break;strcpy (ans, f[i].str);strcat (ans, f[j].str);if (judge (i, j))return;strcpy (ans, f[j].str);strcat (ans, f[i].str);if (judge (i, j))return;}}}int main () {int t;char ch;scanf ("%d%*c", &t);getchar();while (t--) {n = sum = 0;while (gets(f[n].str) != NULL) {f[n].len = strlen (f[n].str);sum += f[n].len;if (strcmp(f[n].str,"") == 0)break;n++;}sum = sum * 2 / n;//printf ("%d %d\n", sum, n);sort (f, f + n, cmp);/*for (int i = 0; i < n; i++)printf ("%s\n", f[i].str);*/solve();printf ("%s\n", ans);if (t)printf ("\n");}return 0;}