[BZOJ 1006] [HNOI2008] 神奇的国度 【弦图最小染色】

题目链接: BZOJ - 1006

题目分析

这道题是一个弦图最小染色数的裸的模型。

弦图的最小染色求法，先求出弦图的完美消除序列（MCS算法），再按照完美消除序列，从后向前倒着，给每个点染能染的最小颜色。

求出的颜色数就是最小染色，同时也是最大团。

代码

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int MaxN = 10000 + 5, MaxM = 1000000 + 5;

int n, m, Ans;
int V[MaxN], A[MaxN], Col[MaxN], Used[MaxN];

bool Visit[MaxN];

struct Edge
{
int v;
Edge *Next;
} E[MaxM * 2], *P = E, *Point[MaxN];

inline void AddEdge(int x, int y)
{
++P; P -> v = y;
P -> Next = Point[x]; Point[x] = P;
}

struct ES
{
int p, q;
ES() {}
ES(int a, int b)
{
p = a; q = b;
}
};

struct Cmp
{
bool operator () (ES e1, ES e2)
{
return e1.q < e2.q;
}
};

priority_queue<ES, vector<ES>, Cmp> Q;

//MCS 求完美消除序列
void MCS()
{
for (int i = 1; i <= n; ++i)
{
V[i] = 0;
Visit[i] = false;
}
while (!Q.empty()) Q.pop();
Q.push(ES(1, 0));
int x, y;
for (int i = n; i >= 1; --i)
{
while (true)
{
x = Q.top().p; Q.pop();
if (!Visit[x]) break;
}
A[i] = x;
Visit[x] = true;
for (Edge *j = Point[x]; j; j = j -> Next)
{
y = j -> v;
if (Visit[y]) continue;
++V[y];
Q.push(ES(y, V[y]));
}
}
}

void Min_Paint()
{
Ans = 1;
int x;
memset(Col, 0, sizeof(Col));
memset(Used, 0, sizeof(Used));
for (int i = n; i >= 1; --i)
{
for (Edge *j = Point[A[i]]; j; j = j -> Next)
Used[Col[j -> v]] = i;
x = 1;
while (Used[x] == i)
{
++x;
if (x > Ans) Ans = x;
}
Col[A[i]] = x;
}
}

int main()
{
scanf("%d%d", &n, &m);
int a, b;
for (int i = 1; i <= m; ++i)
{
scanf("%d%d", &a, &b);
AddEdge(a, b);
AddEdge(b, a);
}
MCS();
Min_Paint();
printf("%d\n", Ans);
return 0;
}