这题看了好久才懂什么意思。
简单地说就是给出一个数字原来的位置,和最后的位置,求移动顺序。
输入簇(cluster)的总数和文件个数,每个文件个数后面跟着的是各个簇的原始位置。最后要按顺序排下来。
分两种情况。
如果没有环,直接DFS然后换掉。
如果有环,从后面开始找到第一个空的位置,然后把数移过去,然后就递归了。
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <cstdlib>
#include <fstream>
#include <string>
#include <sstream>
#include <map>
#include <cmath>
#define LL long long
#define lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 1e4 + 100;
const int MOD = 20071027;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int num[MAXN], vis[MAXN], N;
void DFS(int n)
{
if (num[num[n]] == 0)
{
printf("%d %d\n", n, num[n]);
num[num[n]] = -1;
num[n] = 0;
return;
}
if (vis[num[num[n]]])
{
for (int i = N; i > 0; i--)
{
if (num[i] == 0)
{
printf("%d %d\n", n, i);
num[i] = num[n];
num[n] = 0;
return;
}
}
}
vis[n] = 1;
DFS(num[n]);
printf("%d %d\n", n, num[n]);
num[num[n]] = -1;
num[n] = 0;
}
int main()
{
//ROP;
int T, i, j, nfile, n;
scanf("%d", &T);
while (T--)
{
MS(num, 0);
int pos = 1, tmp;
scanf("%d%d", &N, &nfile);
for (i = 0; i < nfile; i++)
{
scanf("%d", &n);
for (j = 0; j < n; j++)
{
scanf("%d", &tmp);
num[tmp] = pos++;
if (tmp == pos - 1) num[tmp] = -1; //already placed in the correct position
}
}
bool rec = false;
for (i = 1; i <= N; i++)
{
if (num[i] && num[i] != -1)
{
MS(vis, 0);
vis[i] = 1;
DFS(i);
rec = true;
}
}
if (!rec) puts("No optimization needed");
if (T) puts("");
}
return 0;
}