HDU 2236 Untitled II (binary, enumeration, binary graph maximum matching)

Untitled II Time limit:2000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1515 accepted submission (s): 688


Problem Description This is a simple game, in a n*n matrix, find the number of n so that the number of n in a different row and column and require that n number of the maximum and minimum difference is the minimum value.
Input enter an integer t to represent the T-group data.
Enter a positive integer n (1<=n<=100) for the first row of each set of data to represent the size of the matrix.
Then enter n rows, n number x (0<=x<=100) per row.

Output represents the minimum value for each set of data outputs.
Sample Input

1 4 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4-4
Sample Output

3
Author XHD
Source hdoj 2008 Summer Exercise (2)-Hold by Captain Xu: Chinese title does not explain

Idea: Come up and write a DFS, uh ... TLE

Then think of the binary difference, the lower bounds of the enumeration to see whether each row can be matched to (that is, the binary graph maximum matching, behavior x set, column y set). This is the condition of the binary transfer.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <
Cmath> using namespace std;
#define N #define INF 99999999 int N,ans;
int m[n][n],ma[n][n],vis[n];
int num[n],cnt,line[n]; int can (int t) {for (int i=1; i<=n; i++) {if (!vis[i]&&ma[t][i)) {Vis[i]=1
            ; if (line[i]==-1| |
                Can (Line[i])) {line[i]=t;
            return 1;
}} return 0;
    } void get (int t) {for (int i=0;i<cnt;i++) if (num[i]==t) return;
num[cnt++]=t;
    int main () {int T;
    scanf ("%d", &t);
        while (t--) {cnt=0;
        scanf ("%d", &n); for (int i=1; i<=n; i++) for (int j=1; j<=n; J + +) {scanf ("%d", &m[
                    I][J]);
                Get (M[i][j]);
        int l=0,r=210;
        Sort (num,num+cnt);
        int last;while (l<=r) {int mid= (L+R) >>1,flag=0;
                for (int i=0;i<cnt;i++) {memset (ma,0,sizeof (MA)); for (int k=1;k<=n;k++) for (int j=1;j<=n;j++) if (M[K][J]&GT;=NUM[I]&AMP;&A
                Mp;m[k][j]<=num[i]+mid) Ma[k][j]=1;
                memset (line) (line,-1,sizeof);
                int ans=0;
                    for (int j=1;j<=n;j++) {memset (vis,0,sizeof (VIS));
                    if (Can (j)) ans++;
                else break;
                    } if (ans==n) {flag=1;
                Break
            } if (flag) last=mid,r=mid-1;
        else l=mid+1;
    printf ("%d\n", last);
return 0;
 }