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 = 15;
int n;
int G[maxn][maxn];
char s[100];
int sum[maxn], a[maxn];

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

void dfs(int u)
{
    c[u] = 1;
    for(int v = 0; v <= n; v++) if(G[u][v] && !c[v]) dfs(v);
    topo[--t] = u;
}

void toposort()
{
    t = n + 1;
    memset(c, 0, sizeof(c));
    for(int u = 0; 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 = 0; i <= n; i++) p[i] = i;

        memset(G, 0, sizeof(G));
        int p = 0;
        for(int i = 1; i <= n; i++)
            for(int j = i; j <= n; j++)
            {
                if(s[p] == '+') G[i-1][j]++;
                if(s[p] == '-') G[j][i-1]++;
                if(s[p] == '0') link(i-1, j);
                p++;
            }
        toposort();
        for(p = 0; p <= n; p++) if(topo[p] == 0) break;
        s[0] = 0;

        for(int i = p + 1; i <= n; i++)
        {
            int s1 = topo[i], s2 = topo[i-1];
            if(find(s1) == find(s2)) sum[s1] = sum[s2]; //前缀和相等
            else sum[s1] = sum[s2] + 1;
        }
        for(int i = p - 1; i >= 0; i--)
        {
            int s1 = topo[i], s2 = topo[i+1];
            if(find(s1) == find(s2)) sum[s1] = sum[s2];
            else sum[s1] = sum[s2] - 1;
        }
        for(int i = 1; i <= n; i++)
        {
            if(i > 1) printf(" ");
            printf("%d", sum[i] - sum[i - 1]);
        }
        puts("");
    }

    return 0;
}