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

 
  1. #include <cstdio>
  2. #include <stack>
  3. #include <set>
  4. #include <iostream>
  5. #include <string>
  6. #include <vector>
  7. #include <queue>
  8. #include <functional>
  9. #include <cstring>
  10. #include <algorithm>
  11. #include <cctype>
  12. #include <string>
  13. #include <map>
  14. #include <iomanip>
  15. #include <cmath>
  16. #define LL long long
  17. #define ULL unsigned long long
  18. #define SZ(x) (int)x.size()
  19. #define Lowbit(x) ((x) & (-x))
  20. #define MP(a, b) make_pair(a, b)
  21. #define MS(arr, num) memset(arr, num, sizeof(arr))
  22. #define PB push_back
  23. #define F first
  24. #define S second
  25. #define ROP freopen("input.txt", "r", stdin);
  26. #define MID(a, b) (a + ((b - a) >> 1))
  27. #define LC rt << 1, l, mid
  28. #define RC rt << 1|1, mid + 1, r
  29. #define LRT rt << 1
  30. #define RRT rt << 1|1
  31. #define BitCount(x) __builtin_popcount(x)
  32. #define BitCountll(x) __builtin_popcountll(x)
  33. #define LeftPos(x) 32 - __builtin_clz(x) - 1
  34. #define LeftPosll(x) 64 - __builtin_clzll(x) - 1
  35. const double PI = acos(-1.0);
  36. const int INF = 0x3f3f3f3f;
  37. using namespace std;
  38. const double eps = 1e-8;
  39. const int MAXN = 300 + 10;
  40. const int MOD = 1000007;
  41. const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
  42. typedef pair<int, int> pii;
  43. typedef vector<int>::iterator viti;
  44. typedef vector<pii>::iterator vitii;
  46. int mp[MAXN][MAXN];
  47. int pos[MAXN], val[MAXN], n;
  49. void Floyd()
  50. {
  51.     for (int k = 1; k <= n; k++)
  52.         for (int i = 1; i <= n; i++)
  53.             for (int j = 1; j <= n; j++)
  54.                 if (mp[i][j] == 0) mp[i][j] = mp[i][k] & mp[k][j];
  55. }
  57. int main()
  58. {
  59.     //ROP;
  60.     int i, j;
  61.     scanf("%d", &n);
  62.     for (i = 1; i <= n; i++)
  63.     {
  64.         int tmp;
  65.         scanf("%d%*c", &tmp);
  66.         pos[tmp] = i, val[i] = tmp;
  67.     }
  68.     for (i = 1; i <= n; i++)
  69.     {
  70.         for (j = 1; j <= n; j++)
  71.         {
  72.             char tmp;
  73.             scanf("%c", &tmp);
  74.             mp[i][j] = tmp - '0';
  75.         }
  76.         getchar();
  77.     }
  78.     Floyd();
  79.     int conti = 1;
  80.     for (i = 1; i <= n; i++)
  81.         for (j = conti; j <= n; j++)
  82.         {
  83.             if (mp[i][pos[j]])
  84.             {
  85.                 int oriPos = pos[j];
  86.                 val[oriPos] = val[i];
  87.                 pos[val[i]] = oriPos;
  88.                 val[i] = j;
  89.                 pos[j] = i;
  91.             }
  92.         }
  93.     for (i = 1; i <= n; i++)
  94.     {
  95.         if (i == 1) printf("%d", val[i]);
  96.         else printf(" %d", val[i]);
  97.     }
  98.     puts("");
  99.     return 0;
  100. }
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