LA 4255 (拓扑排序 并查集) Guess

设这个序列的前缀和为S_i(0 <= i <= n)，S₀ = 0

每一个符号对应两个前缀和的大小关系，然后根据这个关系拓扑排序一下。

还要注意一下前缀和相等的情况，所以用一个并查集来查询。

 1 #include <cstdio>

 2 #include <cstring>

 3 

 4 const int maxn = 15;

 5 int n;

 6 int G[maxn][maxn];

 7 char s[100];

 8 int sum[maxn], a[maxn];

 9 

10 int topo[maxn], c[maxn], t;

11 

12 void dfs(int u)

13 {

14     c[u] = 1;

15     for(int v = 0; v <= n; v++) if(G[u][v] && !c[v]) dfs(v);

16     topo[--t] = u;

17 }

18 

19 void toposort()

20 {

21     t = n + 1;

22     memset(c, 0, sizeof(c));

23     for(int u = 0; u <= n; u++) if(!c[u]) dfs(u);

24 }

25 

26 int p[maxn];

27 int find(int x)

28 { return x == p[x] ? x : p[x] = find(p[x]); }

29 

30 void link(int x, int y)

31 {

32     int px = find(x), py = find(y);

33     if(px != py) p[px] = py;

34 }

35 

36 int main()

37 {

38     //freopen("in.txt", "r", stdin);

39 

40     int T; scanf("%d", &T);

41     while(T--)

42     {

43         scanf("%d", &n);

44         scanf("%s", s);

45 

46         for(int i = 0; i <= n; i++) p[i] = i;

47 

48         memset(G, 0, sizeof(G));

49         int p = 0;

50         for(int i = 1; i <= n; i++)

51             for(int j = i; j <= n; j++)

52             {

53                 if(s[p] == '+') G[i-1][j]++;

54                 if(s[p] == '-') G[j][i-1]++;

55                 if(s[p] == '0') link(i-1, j);

56                 p++;

57             }

58         toposort();

59         for(p = 0; p <= n; p++) if(topo[p] == 0) break;

60         s[0] = 0;

61 

62         for(int i = p + 1; i <= n; i++)

63         {

64             int s1 = topo[i], s2 = topo[i-1];

65             if(find(s1) == find(s2)) sum[s1] = sum[s2]; //前缀和相等

66             else sum[s1] = sum[s2] + 1;

67         }

68         for(int i = p - 1; i >= 0; i--)

69         {

70             int s1 = topo[i], s2 = topo[i+1];

71             if(find(s1) == find(s2)) sum[s1] = sum[s2];

72             else sum[s1] = sum[s2] - 1;

73         }

74         for(int i = 1; i <= n; i++)

75         {

76             if(i > 1) printf(" ");

77             printf("%d", sum[i] - sum[i - 1]);

78         }

79         puts("");

80     }

81 

82     return 0;

83 }

代码君