1099. Work Scheduling

Time limit: 0.5 second
Memory limit: 64 MB
There is certain amount of night guards that are available to protect the local junkyard from possible junk robberies. These guards need to scheduled in pairs, so that each pair guards at different night. The junkyard CEO ordered you to write a program which given the guards characteristics determines the maximum amount of scheduled guards (the rest will be fired). Please note that each guard can be scheduled with only one of his colleagues and no guard can work alone.

Input

The first line of the input contains one number N ≤ 222 which is the amount of night guards. Unlimited number of lines consisting of unordered pairs (ij) follow, each such pair means that guard #i and guard #j can work together, because it is possible to find uniforms that suit both of them (The junkyard uses different parts of uniforms for different guards i.e. helmets, pants, jackets. It is impossible to put small helmet on a guard with a big head or big shoes on guard with small feet). The input ends with Eof.

Output

You should output one possible optimal assignment. On the first line of the output write the even number C, the amount of scheduled guards. Then output C/2 lines, each containing 2 integers (ij) that denote that i and j will work together.

Sample

input output
3
1 2
2 3
1 3
2
1 2
————————————————————————————————
题目的意思是给出n个士兵和几组关系,士兵两两配对搭档,问最后有多少人有搭档并
输出
思路:一般图匹配模板题,套带花树开花算法模板
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std; #define LL long long
const int INF = 0x3f3f3f3f; const int MAXN = 250;
int N; //点的个数,点的编号从1到N
bool Graph[MAXN][MAXN];
int Match[MAXN];
bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN];
int Head,Tail;
int Queue[MAXN];
int Start,Finish;
int NewBase;
int Father[MAXN],Base[MAXN];
int Count;//匹配数,匹配对数是Count/2 void Push(int u)
{
Queue[Tail] = u;
Tail++;
InQueue[u] = true;
}
int Pop()
{
int res = Queue[Head];
Head++;
return res;
}
int FindCommonAncestor(int u,int v)
{
memset(InPath,false,sizeof(InPath));
while(true)
{
u = Base[u];
InPath[u] = true;
if(u == Start) break;
u = Father[Match[u]];
}
while(true)
{
v = Base[v];
if(InPath[v])break;
v = Father[Match[v]];
}
return v;
}
void ResetTrace(int u)
{
int v;
while(Base[u] != NewBase)
{
v = Match[u];
InBlossom[Base[u]] = InBlossom[Base[v]] = true;
u = Father[v];
if(Base[u] != NewBase) Father[u] = v;
}
}
void BloosomContract(int u,int v)
{
NewBase = FindCommonAncestor(u,v);
memset(InBlossom,false,sizeof(InBlossom));
ResetTrace(u);
ResetTrace(v);
if(Base[u] != NewBase) Father[u] = v;
if(Base[v] != NewBase) Father[v] = u;
for(int tu = 1; tu <= N; tu++)
if(InBlossom[Base[tu]])
{
Base[tu] = NewBase;
if(!InQueue[tu]) Push(tu);
}
}
void FindAugmentingPath()
{
memset(InQueue,false,sizeof(InQueue));
memset(Father,0,sizeof(Father));
for(int i = 1; i <= N; i++)
Base[i] = i;
Head = Tail = 1;
Push(Start);
Finish = 0;
while(Head < Tail)
{
int u = Pop();
for(int v = 1; v <= N; v++)
if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v))
{
if((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0))
BloosomContract(u,v);
else if(Father[v] == 0)
{
Father[v] = u;
if(Match[v] > 0)
Push(Match[v]);
else
{
Finish = v;
return;
}
}
}
}
}
void AugmentPath()
{
int u,v,w;
u = Finish;
while(u > 0)
{
v = Father[u];
w = Match[v];
Match[v] = u;
Match[u] = v;
u = w;
}
}
void Edmonds()
{
memset(Match,0,sizeof(Match));
for(int u = 1; u <= N; u++)
if(Match[u] == 0)
{
Start = u;
FindAugmentingPath();
if(Finish > 0)AugmentPath();
}
} int main()
{ int u,v;
while(~scanf("%d",&N))
{
memset(Graph,false,sizeof(Graph)); while(~scanf("%d%d",&u,&v))
{
Graph[u][v] = Graph[v][u] = true;
} Edmonds();//进行匹配
int cnt=0;
for(int i=1; i<=N; i++)
if(Match[i]>0)
cnt++;
printf("%d\n",cnt);
for(int i=1; i<=N; i++)
if(i<Match[i])
printf("%d %d\n",i,Match[i]); } return 0;
}

  

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