SGU 134 数组模拟树状dp

来源:互联网 发布:朝鲜 氢弹 知乎 编辑:程序博客网 时间:2024/06/06 05:18

往往在dfs层数增加的时候都会面临着爆栈的危险,怎样避免它??栈模拟?不对~栈模拟编程复杂度太高~

用数组模拟!数组模拟不仅快,而且代码量很短~适合搞树状dp~细节多,初学者慎用~

#include<iostream>#include<vector>#include<algorithm>#include<cstdio>#include<queue>#include<stack>#include<string>#include<map>#include<set>#include<cmath>#include<cassert>#include<cstring>#include<iomanip>using namespace std;#ifdef _WIN32#define i64 __int64#define out64 "%I64d\n"#define in64 "%I64d"#else#define i64 long long#define out64 "%lld\n"#define in64 "%lld"#endif#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)#define FFD(i,a)        for( int i = (a)-1 ; i >= 0 ; i --)#define S64(a)          scanf(in64,&a)#define SS(a)           scanf("%d",&a)#define LL(a)           ((a)<<1)#define RR(a)           (((a)<<1)+1)#define SZ(a)           ((int)a.size())#define PP(n,m,a)       puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}#define pb              push_back#define CL(Q)           while(!Q.empty())Q.pop()#define MM(name,what)   memset(name,what,sizeof(name))#define read            freopen("in.txt","r",stdin)#define write           freopen("out.txt","w",stdout)const int inf = 0x3f3f3f3f;const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;const double oo = 10e9;const double eps = 10e-10;const double pi = acos(-1.0);const int maxn = 16001;vector<int>g[maxn];vector<int>v;int dp[maxn];int dpmax[maxn];int f[maxn];int dfn[maxn];  int dfv[maxn];int df,n,ans;void dfs(int now=1){    FF(i,g[now].size()) if(g[now][i] != f[now])    {        f[g[now][i]] = now;        dfs(g[now][i]);    }    dfv[df] = now;    dfn[now] = df;    df++;    return ;}void dps(){    FOR(i,1,df-2)    // 这个地方小心,-1的话就是wa,因为根节点没有父节点    {        dp[dfv[i]]++;        dpmax[dfv[i]]=max(dpmax[dfv[i]],n-dp[dfv[i]]);        dpmax[f[dfv[i]]]=max(dpmax[f[dfv[i]]],dp[dfv[i]]);        dp[f[dfv[i]]]+=dp[dfv[i]];    }    return ;}void start(){    df = 1;    MM(dp,0);    MM(dpmax,0);    MM(f,0);    dfs();    dps();    ans=inf;    for(int i=1;i<=n;i++)    {        ans = min(ans,dpmax[i]);    }    for(int i=1;i<=n;i++)    {        if(dpmax[i] == ans)        {            v.pb(i);        }    }    sort(v.begin(),v.end());    return ;}int main(){    cin>>n;    int now,to;    FOR(i,1,n-1)    {        cin>>now>>to;        g[now].pb(to);        g[to].pb(now);    }    start();    cout<<ans<<" "<<v.size()<<endl;    FF(i,v.size()-1)    {        cout<<v[i]<<" ";    }    cout<<v.back()<<endl;    return 0;}


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