URAL 1099. Work Scheduling 一般图匹配带花树
2014-11-05 14:51
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一般图匹配带花树模版题:
将奇环缩成圈(Blossom),然后找增广路.....
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.
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.
integers (i, j) that denote that i and j will work together.
Problem Author: Jivko Ganev
Tags: graph theory (
hide tags for unsolved problems
)
Difficulty: 1356 Printable version Submit solution Discussion
(51)
My submissions All submissions (15804) All
accepted submissions (1687) Solutions rating (610)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=250;
/*******************************************/
struct Edge
{
int to,next;
}edge[maxn*maxn];
int Adj[maxn],Size;
void init()
{
memset(Adj,-1,sizeof(Adj)); Size=0;
}
void add_edge(int u,int v)
{
edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}
/*******************************************/
int n;
int Match[maxn];
bool G[maxn][maxn];
int Start,Finish,NewBase;
int Father[maxn],Base[maxn];
bool InQueue[maxn],InPath[maxn],InBlossom[maxn];
int Count;
queue<int> q;
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 BlosomContract(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])
{
q.push(tu);
InQueue[tu]=true;
}
}
}
}
void FindAugmentingPath()
{
memset(InQueue,false,sizeof(InQueue));
memset(Father,0,sizeof(Father));
for(int i=1;i<=n;i++)
Base[i]=i;
while(!q.empty()) q.pop();
q.push(Start); InQueue[Start]=true;
Finish=0;
while(!q.empty())
{
int u=q.front(); //InQueue[u]=false;
q.pop();
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if((Base[u]!=Base[v])&&Match[u]!=v)
{
if(v==Start||(Match[v]>0&&Father[Match[v]]>0))
BlosomContract(u,v);
else if(Father[v]==0)
{
Father[v]=u;
if(Match[v]>0)
{
q.push(Match[v]);
InQueue[Match[v]]=true;
}
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();
}
}
}
void PrintMatch()
{
Count=0;
for(int i=1;i<=n;i++)
{
if(Match[i]) Count++;
}
printf("%d\n",Count);
for(int u=1;u<=n;u++)
{
if(u<Match[u])
printf("%d %d\n",u,Match[u]);
}
}
int main()
{
//freopen("data.in","r",stdin);
while(scanf("%d",&n)!=EOF)
{
init();
memset(G,false,sizeof(G));
int u,v;
while(scanf("%d%d",&u,&v)!=EOF)
{
if(G[u][v]==true) continue;
G[u][v]=G[v][u]=true;
add_edge(u,v);
add_edge(v,u);
}
Edmonds();
PrintMatch();
}
return 0;
}
将奇环缩成圈(Blossom),然后找增广路.....
1099. Work Scheduling
Time limit: 0.5 secondMemory 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 (i, j) follow, each suchpair 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 2integers (i, j) that denote that i and j will work together.
Sample
input | output |
---|---|
3 1 2 2 3 1 3 | 2 1 2 |
Tags: graph theory (
hide tags for unsolved problems
)
Difficulty: 1356 Printable version Submit solution Discussion
(51)
My submissions All submissions (15804) All
accepted submissions (1687) Solutions rating (610)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=250;
/*******************************************/
struct Edge
{
int to,next;
}edge[maxn*maxn];
int Adj[maxn],Size;
void init()
{
memset(Adj,-1,sizeof(Adj)); Size=0;
}
void add_edge(int u,int v)
{
edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}
/*******************************************/
int n;
int Match[maxn];
bool G[maxn][maxn];
int Start,Finish,NewBase;
int Father[maxn],Base[maxn];
bool InQueue[maxn],InPath[maxn],InBlossom[maxn];
int Count;
queue<int> q;
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 BlosomContract(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])
{
q.push(tu);
InQueue[tu]=true;
}
}
}
}
void FindAugmentingPath()
{
memset(InQueue,false,sizeof(InQueue));
memset(Father,0,sizeof(Father));
for(int i=1;i<=n;i++)
Base[i]=i;
while(!q.empty()) q.pop();
q.push(Start); InQueue[Start]=true;
Finish=0;
while(!q.empty())
{
int u=q.front(); //InQueue[u]=false;
q.pop();
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if((Base[u]!=Base[v])&&Match[u]!=v)
{
if(v==Start||(Match[v]>0&&Father[Match[v]]>0))
BlosomContract(u,v);
else if(Father[v]==0)
{
Father[v]=u;
if(Match[v]>0)
{
q.push(Match[v]);
InQueue[Match[v]]=true;
}
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();
}
}
}
void PrintMatch()
{
Count=0;
for(int i=1;i<=n;i++)
{
if(Match[i]) Count++;
}
printf("%d\n",Count);
for(int u=1;u<=n;u++)
{
if(u<Match[u])
printf("%d %d\n",u,Match[u]);
}
}
int main()
{
//freopen("data.in","r",stdin);
while(scanf("%d",&n)!=EOF)
{
init();
memset(G,false,sizeof(G));
int u,v;
while(scanf("%d%d",&u,&v)!=EOF)
{
if(G[u][v]==true) continue;
G[u][v]=G[v][u]=true;
add_edge(u,v);
add_edge(v,u);
}
Edmonds();
PrintMatch();
}
return 0;
}
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