bzoj 1006: [HNOI2008]神奇的国度

这是个标准的弦图，但如果不知道弦图就惨了=_=

趁着这个机会了解了一下弦图，主要就是完美消除序列，求出了这个就可以根据序列进行贪心染色。

貌似这个序列很神，但是具体应用不了解……

这道题为什么可以这么做不理解……

我真是太弱了……

上代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <queue>
#define N 100100
#define M 2000100
using namespace std;
struct sss
{
    int num;
    int du;
};
int n, m;
int p[N] = {}, next[M], v[M], bnum = ;
int du[N] = {}, vis[N] = {}, qu[N];
priority_queue<sss> q;
int hcolor[N] = {}, color[N] = {}, colornum = , huse[N] = {};

bool operator < (sss x, sss y)
{
    return x.du < y.du;
}

void addbian(int x, int y)
{
    bnum++; next[bnum] = p[x]; p[x] = bnum; v[bnum] = y;
    bnum++; next[bnum] = p[y]; p[y] = bnum; v[bnum] = x;
}

void make_queue()
{
    for (int i = ; i <= n; ++i)
    {
        sss x; x.num = i; x.du = ;
        q.push(x);
    }
    int dc = ;
    while (dc < n)
    {
        sss y = q.top(); q.pop();
        while (vis[y.num])
        {
            y = q.top();
            q.pop();
        }
        vis[y.num] = ;
        qu[++dc] = y.num;
        int k = p[y.num]; sss ne;
        while (k)
        {
            if (!vis[v[k]])
            {
                du[v[k]]++; ne.du = du[v[k]];
                ne.num = v[k]; q.push(ne);
            }
            k = next[k];
        }
    }
}

void color_dian()
{
    for (int i = ; i <= n; ++i)
    {
        int x = qu[i];
        int k = p[x];
        int all = ;
        while (k)
        {
            if (color[v[k]] != ) all = ;
            huse[color[v[k]]] = x;
            k = next[k];
        }
        int pd = ;
        for (int i = ; i <= colornum; ++i)
            if (huse[i] != x)
            {
                pd = ;
                color[x] = i;
                break;
            }
        if (!pd) color[x] = ++colornum;
    }
    printf("%d\n", colornum);
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = ; i <= m; ++i)
    {
        int x, y; scanf("%d%d", &x, &y);
        addbian(x, y);
    }
    make_queue();
    color_dian();
    return ;
}