BZOJ1006 [HNOI2008]神奇的国度
标签: 弦图
有关弦图的定义与mcs算法请自学(模板)
题意为求弦图的色数\(\chi(G)\)
做法:
最优染色构造:将完美消除序列从后往前依次给每个点染色,给每个点染上可染的最小颜色
证明:
设以上方法用\(col\)个颜色,原图团数为\(\omega(G)\)
引理:\(\omega(G) \leq \chi(G)\)
引理证明:对最大团的导出子图染色,至少需要\(\omega(G)\)种颜色(因为团中每个点都相邻,每个点都需用不同的颜色)
设\(col \geq \chi(G)\)
团上每个点都不同\(\rightarrow col = \omega(G)\)
由引理:\(col = \omega(G) \leq \chi(G)\),\(col \geq \chi(G)\)
$\therefore t = \omega(G) =\chi(G) $
\(ans = max\{|\{x\} + N(x)|\} = max\{lab_i\} + 1\)
代码: 要开O2优化
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1e4 + 1;
int n, m, u, v, lab[N], p[N];
bool c[N][N], vis[N];
inline void mcs()
{
for (int i = n; i >= 1; --i)
{
u = 0;
for (int j = 1; j <= n; ++j)
if (!vis[j] && (!u || lab[j] > lab[u]))
u = j;
vis[u] = 1;
p[i] = u;
for (int j = 1; j <= n; ++j)
if (!vis[j] && c[u][j])
++lab[j];
}
}
int main()
{
ios::sync_with_stdio(false);
cin >> n >> m;
for (int i = 1; i <= m; ++i)
{
cin >> u >> v;
c[u][v] = c[v][u] = 1;
}
mcs();
int ans = 0;
for (int i = 1; i <= n; ++i)
ans = max(ans, lab[i] + 1);
cout << ans << endl;
return 0;
}