hdoj 3861 The King’s Problem【强连通缩点建图&&最小路径覆盖】
The King’s Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2259 Accepted Submission(s):
795
There are N cities in the kingdom and there are M directional roads between the
cities. That means that if there is a road from u to v, you can only go from
city u to city v, but can’t go from city v to city u. In order to rule his
kingdom more effectively, the king want to divide his kingdom into several
states, and each city must belong to exactly one state. What’s
more, for each pair of city (u, v), if there is one way to go from u to v and go
from v to u, (u, v) have to belong to a same state. And the king must
insure that in each state we can ether go from u to v or go from v to u between
every pair of cities (u, v) without passing any city which belongs to other
state.
Now the king asks for your help, he wants to know the least number
of states he have to divide the kingdom into.
of test cases. And then followed T cases.
The first line for each case
contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the
number of cities and roads in the kingdom. The next m lines each contains two
integers u and v (1 <= u, v <= n), indicating that there is a road going
from city u to city v.
you should just output an integer which is the least number of states the king
have to divide into.
#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#define MAX 5200
#define MAXM 200100
using namespace std;
vector<int>newmap[MAX];
vector<int>scc[MAX];
int sccno[MAX];
int in[MAX],out[MAX];
int scccnt,dfsclock;
int n,m;
int low[MAX],dfn[MAX];
int instack[MAX];
int ans,head[MAX];
int vis[MAX],city[MAX];
stack<int>s;
struct node
{
int beg,end,next;
}edge[MAXM];
void init()
{
ans=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
edge[ans].beg=u;
edge[ans].end=v;
edge[ans].next=head[u];
head[u]=ans++;
}
void getmap()
{
int a,b;
while(m--)
{
scanf("%d%d",&a,&b);
add(a,b);
}
}
void tarjan(int u)
{
int v,i,j;
low[u]=dfn[u]=++dfsclock;
s.push(u);
instack[u]=1;
for(i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].end;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(instack[v])
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
scccnt++;
while(1)
{
v=s.top();
s.pop();
instack[v]=0;
sccno[v]=scccnt;
if(v==u)
break;
}
}
}
void find(int l,int r)
{
memset(low,0,sizeof(low));
memset(dfn,0,sizeof(dfn));
memset(instack,0,sizeof(instack));
memset(sccno,0,sizeof(sccno));
dfsclock=scccnt=0;
for(int i=l;i<=r;i++)
{
if(!dfn[i])
tarjan(i);
}
}
void suodian()
{
find(1,n);
for(int i=1;i<=scccnt;i++)
newmap[i].clear();
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
int u,v,i,j;
for(i=0;i<ans;i++)
{
u=sccno[edge[i].beg];
v=sccno[edge[i].end];
if(u!=v)
{
newmap[u].push_back(v);
in[v]++;
out[u]++;
}
}
}
int query(int x)
{
int i,j;
for(i=0;i<newmap[x].size();i++)
{
int y=newmap[x][i];
if(!vis[y])
{
vis[y]=1;
if(city[y]==0||query(city[y]))
{
city[y]=x;
return 1;
}
}
}
return 0;
}
void solve()
{
int i,j;
int sum=0;
memset(city,0,sizeof(city));
for(i=1;i<=scccnt;i++)
{
memset(vis,0,sizeof(vis));
if(query(i))
sum++;
}
printf("%d\n",scccnt-sum);//最小路径覆盖=顶点数-最大匹配数
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
getmap();
suodian();
solve();
}
return 0;
}
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