1006: [HNOI2008]神奇的國度
Time Limit: 20 Sec Memory Limit: 162 MBSubmit: 2463 Solved: 1110
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Description
K國是一個熱衷三角形的國度,連人的交往也只喜歡三角原則.他們認爲三角關係:即AB相互認識,BC相互認識,CA相互認識,是簡潔高效的.爲了鞏固三角關係,K國禁止四邊關係,五邊關係等等的存在.所謂N邊關係,是指N個人 A1A2...An之間僅存在N對認識關係:(A1A2)(A2A3)...(AnA1),而沒有其它認識關係.比如四邊關係指ABCD四個人 AB,BC,CD,DA相互認識,而AC,BD不認識.全民比賽時,爲了防止做弊,規定任意一對相互認識的人不得在一隊,國王相知道,最少可以分多少支隊。
Input
第一行兩個整數N,M。1<=N<=10000,1<=M<=1000000.表示有N個人,M對認識關係. 接下來M行每行輸入一對朋友
Output
輸出一個整數,最少可以分多少隊
Sample Input
1 2
1 4
2 4
2 3
3 4
Sample Output
這種松蘑幼兒園就做過的題也沒啥好講的,就是好久沒更新blog了來水一水,題解詳見CDQ《弦圖與區間圖》
代碼如下:
/**************************************************************
Problem: 1006
User: duyixian
Language: C++
Result: Accepted
Time:1064 ms
Memory:29424 kb
****************************************************************/
/*
* @Author: duyixian
* @Date: 2015-09-08 16:02:39
* @Last Modified by: duyixian
* @Last Modified time: 2015-09-08 16:46:27
*/
#include "cstdio"
#include "cstdlib"
#include "iostream"
#include "algorithm"
#include "cstring"
#include "queue"
using namespace std;
#define MAX_SIZE 10005
#define INF 0x3F3F3F3F
#define Eps
#define Mod
inline int Get_Int()
{
int Num = 0, Flag = 1;
char ch;
do
{
ch = getchar();
if(ch == '-')
Flag *= -1;
}
while(ch < '0' || ch > '9');
do
{
Num = Num * 10 + ch - '0';
ch = getchar();
}
while(ch >= '0' && ch <= '9');
return Num *Flag;
}
struct Edge
{
int To, Next;
}Edges[2000005];
int N, M, Total;
int Front[MAX_SIZE], Sequence[MAX_SIZE], Label[MAX_SIZE], Order[MAX_SIZE], Mark[MAX_SIZE], Color[MAX_SIZE];
priority_queue< pair<int, int> > Heap;
inline void Add_Edge(int From, int To)
{
Edges[++Total].To = To;
Edges[Total].Next = Front[From];
Front[From] = Total;
}
inline void Add_Edges(int From, int To)
{
Add_Edge(From, To);
Add_Edge(To, From);
}
inline void Find()
{
for(int i = 1; i <= N; ++i)
Heap.push(make_pair(0, i));
for(int j = N; j >= 1; )
{
int Now = Heap.top().second;
Heap.pop();
if(Order[Now])
continue;
Sequence[j] = Now;
Order[Now] = j--;
for(int i = Front[Now]; i; i = Edges[i].Next)
{
if(Order[Edges[i].To])
continue;
++Label[Edges[i].To];
Heap.push(make_pair(Label[Edges[i].To], Edges[i].To));
}
}
}
inline void Coloring(int Now)
{
for(int i = Front[Now]; i; i = Edges[i].Next)
{
if(!Color[Edges[i].To])
continue;
Mark[Color[Edges[i].To]] = Now;
}
for(int i = 1; i <= N; ++i)
if(Mark[i] != Now)
{
Color[Now] = i;
break;
}
}
int main()
{
cin >> N >> M;
for(int i = 1; i <= M; ++i)
Add_Edges(Get_Int(), Get_Int());
Find();
for(int i = N; i >= 1; --i)
Coloring(Sequence[i]);
int Ans = 0;
for(int i = 1; i <= N; ++i)
Ans = max(Ans, Color[i]);
cout << Ans << endl;
return 0;
}