2015
09-17

# Backup Plan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 136    Accepted Submission(s): 54
Special Judge

Problem Description
Makomuno has N servers and M databases. All databases are synchronized among all servers and each database has a ordered list denotes the priority of servers to access. This list is guaranteed to be a valid permutation of all servers.
Every time someone wants to execute queries on a certain database, he will send a request to the first server in the list. If it’s dead, he will simply turn to the next one. Otherwise a working copy of the database is found, and this copy is called active.
Now, given N and M, Makomuno wants to find a permutation for each database which could assure that all servers are load-balanced. Moreover, Makomuno hopes the system will be load-balanced even if exactly one server is broken.
Note that if we call the number of active copies on i-th server Ai, then load-balanced means max∣Ai - Aj∣≤1 for any i and j in non broken servers set. We won’t consider broken servers in this case.

Input
The input contains several test cases, terminated by EOF.
Each test case has one line containing two integer N ( 2≤N≤100) and M ( 1≤M≤100).

Output
For each case output M lines, the i-th line contains a permutation of all servers, indicating the expected order. Servers are numbered from 1 to n.

Sample Input
5 3

Sample Output
2 4 3 1 5
1 5 4 2 3
3 5 2 4 1
Hint
In the sample test case, the active copies of these databases are on server 2,1 and 3 in normal state. A = {1,1,1,0,0}
If server 1 or 3 has broken, server 5 will take its work. In case we lost server 2, the second database will use server 4 instead. A = {1,BROKEN,1,1,0}
It's clear that in any case this system is load-balanced according to the plan in sample output.


Source

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zhuyuanchen520

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int V = 105;
int n, m, ans[V][V];
bool vis[V][V];
int main() {
int i, j, k;
while(~scanf("%d%d", &n, &m)) {
memset(ans, 0, sizeof(ans));
memset(vis, false, sizeof(vis));
for(i = 0, j = 1; i < m; ++i) {
ans[i][0] = j;
vis[i][j] = true;
j = j == n ? 1 : j + 1;
}
for(i = 0; i < m; ++i) {
if(ans[i][1] != 0)
continue;
int now = m % n + 1;
for(j = i; j < m; j += n) {
if(now == ans[j][0])
now = now == n ? 1 : now + 1;
ans[j][1] = now;
vis[j][now] = true;
now = now == n ? 1 : now + 1;
}
}
for(i = 0; i < m; ++i)
for(j = 2, k = 1; j < n; ++j) {
while(vis[i][k])
k++;
ans[i][j] = k++;
}
for(i = 0; i < m; ++i) {
for(j = 0; j < n - 1; ++j)
printf("%d ", ans[i][j]);
printf("%d\n", ans[i][n - 1]);
}
}
}