二次联通门 : 

 

 

 

 

/*    BZOJ 1006: [HNOI2008]神奇的国度    弦图    然而SB的我就是看不懂为什么要这样做QAQ    大体做法就是    从大到小扫描每个点，找出与这个点相连的度数最大的点，扔进队列里    然后将这个度数最大的点作为当前点，将于此点相连的点度数都加1    从后往前扫描队列里的元素，进行染色，每个点都染上最小的颜色就好了。*/#include 
     
      #include 
      
       const int BUF = 12312323; char Buf[BUF], *buf = Buf;#define rg register#define Max 10009inline void read (int &n){    for (n = 0; !isdigit (*buf); ++ buf);    for (; isdigit (*buf); n = n * 10 + *buf - '0', ++ buf);}inline void cmax (int &a, int b) { if (b > a) a = b; }struct E { E *n; int v; } *list[Max], poor[Max * 300], *Ta = poor;bool is[Max]; int d[Max], q[Max], k[Max], c[Max];int main (int argc, char *argv[]){    fread (buf, 1, BUF, stdin); E *e; int x, y, Answer = 0;    int N, M, p, n; read (N), read (M); rg int i, j;     for (i = 1, d[0] = -1; i <= M; ++ i)     {        read (x), read (y);        ++ Ta, Ta->v = y, Ta->n = list[x], list[x] = Ta;        ++ Ta, Ta->v = x, Ta->n = list[y], list[y] = Ta;    }    for (i = N; i >= 1; -- i)    {        for (j = 1, n = 0; j <= N; ++ j)            if (!is[j] && d[j] > d[n]) n = j;        for (is[n] = true, q[i] = n, e = list[n]; e; ++ d[e->v], e = e->n);    }        for (i = N; i >= 1; -- i)    {        for (e = list[q[i]]; e; k[c[e->v]] = i, e = e->n);        for (j = 1; k[j] == i; ++ j); c[q[i]] = j; cmax (Answer, j);    }    printf ("%d", Answer);    return 0;}