Codeforces 500B - New Year Permutation (思维)

题意

要得到尽量从小到大排列的序列，但是只有mp[i][j] = 1的时候才能交换两个位置。问能得到的最小序列。

思路

显然应该把小的尽量往前换，我是先对图跑一遍Floyd，然后对每个位置从小到大扫描一遍，能换就换。hcbbt巨巨用了并查集，时间比我少一半Orz。

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cctype>
   
   
   
   
#include <string>
   
   
   
   
#include <map>
   
   
   
   
#include <iomanip>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define ULL unsigned long long
   
   
   
   
#define SZ(x) (int)x.size()
   
   
   
   
#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 MID(a, b) (a + ((b - a) >> 1))
   
   
   
   
#define LC rt << 1, l, mid
   
   
   
   
#define RC rt << 1|1, mid + 1, r
   
   
   
   
#define LRT rt << 1
   
   
   
   
#define RRT rt << 1|1
   
   
   
   
#define BitCount(x) __builtin_popcount(x)
   
   
   
   
#define BitCountll(x) __builtin_popcountll(x)
   
   
   
   
#define LeftPos(x) 32 - __builtin_clz(x) - 1
   
   
   
   
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const double eps = 1e-8;
   
   
   
   
const int MAXN = 300 + 10;
   
   
   
   
const int MOD = 1000007;
   
   
   
   
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
int mp[MAXN][MAXN];
   
   
   
   
int pos[MAXN], val[MAXN], n;
   
   
   
   

   
   
   
   
void Floyd()
   
   
   
   
{
   
   
   
   
 for (int k = 1; k <= n; k++)
   
   
   
   
 for (int i = 1; i <= n; i++)
   
   
   
   
 for (int j = 1; j <= n; j++)
   
   
   
   
 if (mp[i][j] == 0) mp[i][j] = mp[i][k] & mp[k][j];
   
   
   
   
}
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 int i, j;
   
   
   
   
 scanf("%d", &n);
   
   
   
   
 for (i = 1; i <= n; i++)
   
   
   
   
 {
   
   
   
   
 int tmp;
   
   
   
   
 scanf("%d%*c", &tmp);
   
   
   
   
 pos[tmp] = i, val[i] = tmp;
   
   
   
   
 }
   
   
   
   
 for (i = 1; i <= n; i++)
   
   
   
   
 {
   
   
   
   
 for (j = 1; j <= n; j++)
   
   
   
   
 {
   
   
   
   
 char tmp;
   
   
   
   
 scanf("%c", &tmp);
   
   
   
   
 mp[i][j] = tmp - '0';
   
   
   
   
 }
   
   
   
   
 getchar();
   
   
   
   
 }
   
   
   
   
 Floyd();
   
   
   
   
 int conti = 1;
   
   
   
   
 for (i = 1; i <= n; i++)
   
   
   
   
 for (j = conti; j <= n; j++)
   
   
   
   
 {
   
   
   
   
 if (mp[i][pos[j]])
   
   
   
   
 {
   
   
   
   
 int oriPos = pos[j];
   
   
   
   
 val[oriPos] = val[i];
   
   
   
   
 pos[val[i]] = oriPos;
   
   
   
   
 val[i] = j;
   
   
   
   
 pos[j] = i;
   
   
   
   

   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 for (i = 1; i <= n; i++)
   
   
   
   
 {
   
   
   
   
 if (i == 1) printf("%d", val[i]);
   
   
   
   
 else printf(" %d", val[i]);
   
   
   
   
 }
   
   
   
   
 puts("");
   
   
   
   
 return 0;
   
   
   
   
}