Investigating Legions

链接：https://ac.nowcoder.com/acm/contest/5669/I
来源：牛客网

ZYB likes to read detective novels and detective cartoons. One day he is immersed in such story:

As a spy of country B, you have been lurking in country A for many years. There are $n$ troops and $m$ legions in country A, and each troop belongs to only one legion. Your task is to investigate which legion each troop belongs to.

One day, you intercept a top secret message–a paper which records all the pairwise relationship between troops. You can learn from the paper that whether troop $i$ and troop $j$ belongs to the same legion for each $(i,j)$. However, each piece of data has a independent probability of $\frac{1}{S}$ being reversed.
Formally, the troop is numbered from $0$ to $n-1$, and the region is numbered from $0$ to $m-1$. You are given $\frac{n(n-1)}{2}$ boolean numbers, indicating whether $i$ and $j$ belong to the same region ($1$ for yes and $0$ for no). However, every number will be reversed(i.e., $0$ to $1$ and $1$ to $0$) in a probablity of $\frac{1}{S}$. You can assume that the data is generated like this:

$be[i]$ means the region that troop $i$ belongs to. $rand()$ can be considered as a uniform random function. i.e., select a integer from interval $[0,\mathrm{l}\mathrm{c}\mathrm{m}(m,S)-1]$ with equal probability.

Note that you have already known $n$ and $S$, but you don’t know the exact value of $m$. Please help country B construct the original data.
输入描述:
The input contains multiple cases. The first line of the input contains a single integer $T(1\le T\le 100)$, the number of cases.

For each case, the first line of the input contains two integers $n,S$ ($30\le n\le 300,20\le S\le 100$), denoting the number of troops and the integer parameter to generate data. There is a binary string (i.e. the string only contains ‘0’ and ‘1’) with length $\frac{n(n-1)}{2}$ in the next line, decribing the pairwise relationship. It is arrayed like this: $(0,1),(0,2)...(0,n-1),(1,2)\dots (n-3,n-1),(n-2,n-1)$.

$m$ is not given but it is guaranteed that $1\le m\le \lfloor \frac{n}{30}\rfloor$. And the sum of $n$ over $T$ cases doesn’t exceed $1200$.
输出描述:
For each case, output integers seperated by space, the $i$-th integer describes $be[i]$. The value of $be[i]$ is meaningless, so please output the solution with the smallest lexicographical order. For example, if there are $4$ troops and $2$ regions $\{0,3\},\{1,2\}$, you should output $0110$.
示例1

输入
1
10 20
101110101010101010100010010101010100101010010
输出
0 0 1 0 1 0 1 0 1 0

备注:
The sample input does not follow the input format, and it won’t appear in the final test. The parameter is $n=10,m=2,S=20$.
每次选择一个未被编号的点$i$找出所有与$i$相连且未被编号的点，遍历所有点。对于当前遍历到的点$j$，如果$j$看未被标记，则统计与$i$直接相连的点中有多少与$j$相连。显然相连的点数越多越能说明$i$和$j$属于同一类。由于未知变量太多，不能严格证明，只能通过玄学调出判定的边界。

#include 

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scanf("%c",&a)
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector
#define pii pair
#define mii unordered_map
#define msi unordered_map
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 305;
bool g[N][N];
int a[N], n, s, T;
char c;
vi tmp;

int main() {
    T = qr();
    while (T--) {
        n = qr(), s = qr();
        memset(a, -1, sizeof(a));
        repi(i, 1, n - 1) {
            repi(j, i + 1, n)sc(c), g[i][j] = g[j][i] = c - '0';
            g[i][i] = true;
        }
        int id = 0;
        repi(i, 1, n)if (a[i] == -1) {
                tmp.clear();
                repi(j, 1, n)if (g[i][j] && a[j] == -1)tmp.pb(j);
                repi(j, 1, n)
                    if (a[j] == -1) {
                        int cnt = 0;
                        for (auto k:tmp)if (g[j][k])cnt++;
                        if (cnt > tmp.size() / 2)a[j] = id;
                    }
                id++;
            }
        repi(i, 1, n)printf("%d%c", a[i], ce(i, n));
    }
    return 0;
}