Codeforces #264 (Div. 2) D. Gargari and Permutations(DAG求最长路)
2014-09-19 10:55
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思路2:如果一个数字i在每个串的位置都在j前面,那么i到j就有一条有向边,那么题目就转换为DAG求最长。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
int dis[1005],num[5][1005],indu[1005];
int n,k;
vector<int> G[1005];
void init()
{
for(int i=1;i<=n;i++) G[i].clear();
memset(indu,0,sizeof(indu));
memset(dis,0,sizeof(dis));
}
void solve()
{
queue<int> q;
for(int i=1;i<=n;i++)
if(indu[i]==0) dis[i]=1,q.push(i);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(dis[v]<dis[u]+1) dis[v]=dis[u]+1;
if(--indu[v]==0) q.push(v);
}
}
int ans=-1;
for(int i=1;i<=n;i++) ans=max(ans,dis[i]);
cout<<ans<<endl;
}
int main()
{
int a;
cin>>n>>k;
init();
for(int i=0;i<k;i++)
{
for(int j=0;j<n;j++)
{
cin>>a;
num[i][a]=j;
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if(i==j) continue;
int a1=0,b1=0;
for(int t=0;t<k;t++)
if(num[t][i]<num[t][j]) a1++;
else b1++;
if(a1==k) G[i].push_back(j),indu[j]++;
else if(b1==k) G[j].push_back(i),indu[i]++;
}
solve();
return 0;
}
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
int dis[1005],num[5][1005],indu[1005];
int n,k;
vector<int> G[1005];
void init()
{
for(int i=1;i<=n;i++) G[i].clear();
memset(indu,0,sizeof(indu));
memset(dis,0,sizeof(dis));
}
void solve()
{
queue<int> q;
for(int i=1;i<=n;i++)
if(indu[i]==0) dis[i]=1,q.push(i);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(dis[v]<dis[u]+1) dis[v]=dis[u]+1;
if(--indu[v]==0) q.push(v);
}
}
int ans=-1;
for(int i=1;i<=n;i++) ans=max(ans,dis[i]);
cout<<ans<<endl;
}
int main()
{
int a;
cin>>n>>k;
init();
for(int i=0;i<k;i++)
{
for(int j=0;j<n;j++)
{
cin>>a;
num[i][a]=j;
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if(i==j) continue;
int a1=0,b1=0;
for(int t=0;t<k;t++)
if(num[t][i]<num[t][j]) a1++;
else b1++;
if(a1==k) G[i].push_back(j),indu[j]++;
else if(b1==k) G[j].push_back(i),indu[i]++;
}
solve();
return 0;
}
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