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。

 #include <cstdio>
 #include <cstring>

 const int maxn = ;
 int n;
 int G[maxn][maxn];
 char s[];
 int sum[maxn], a[maxn];

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

 void dfs(int u)
 {
     c[u] = ;
     for(int v = ; v <= n; v++) if(G[u][v] && !c[v]) dfs(v);
     topo[--t] = u;
 }

 void toposort()
 {
     t = n + ;
     memset(c, , sizeof(c));
     for(int u = ; u <= n; u++) if(!c[u]) dfs(u);
 }

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

 void link(int x, int y)
 {
     int px = find(x), py = find(y);
     if(px != py) p[px] = py;
 }

 int main()
 {
     //freopen("in.txt", "r", stdin);

     int T; scanf("%d", &T);
     while(T--)
     {
         scanf("%d", &n);
         scanf("%s", s);

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

         memset(G, , sizeof(G));
         int p = ;
         for(int i = ; i <= n; i++)
             for(int j = i; j <= n; j++)
             {
                 if(s[p] == '+') G[i-][j]++;
                 if(s[p] == '-') G[j][i-]++;
                 if(s[p] == '') link(i-, j);
                 p++;
             }
         toposort();
         for(p = ; p <= n; p++) if(topo[p] == ) break;
         s[] = ;

         for(int i = p + ; i <= n; i++)
         {
             int s1 = topo[i], s2 = topo[i-];
             if(find(s1) == find(s2)) sum[s1] = sum[s2]; //前缀和相等
             else sum[s1] = sum[s2] + ;
         }
         for(int i = p - ; i >= ; i--)
         {
             int s1 = topo[i], s2 = topo[i+];
             if(find(s1) == find(s2)) sum[s1] = sum[s2];
             else sum[s1] = sum[s2] - ;
         }
         for(int i = ; i <= n; i++)
         {
             if(i > ) printf(" ");
             printf("%d", sum[i] - sum[i - ]);
         }
         puts("");
     }

     return ;
 }

代码君