10.12NOIP模拟题(2)

/*
有谁知道这道题结论是怎么来的？
晚上问问学数学的孩子23333
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>

using namespace std;
typedef long long LL;
LL n,m,d;

LL gcd(LL x,LL y)
{
    return x%y?gcd(y,x%y):y;
}
int main()
{
    freopen("line.in","r",stdin);
    freopen("line.out","w",stdout);
    cin>>n>>m;
    d=gcd(n,m);
    n/=d;
    m/=d;
    if((n+m)&)cout<<"1/2"<<endl;
    else cout<<(n*m+)/<<"/"<<n*m<<endl;
    return ;
}

/*
数据水...
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100007

using namespace std;
int n,m,opt,x,y,z,ans,cnt;
int col[N];

inline int read()
{
    int x=,f=;char c=getchar();
    while(c>''||c<''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

inline void solve(int l,int r,int flag)
{
    for(int i=l;i<=r;i++) col[i]=flag;
}

inline void find(int x)
{
    int L=,R=n;
    if(col[x]==) printf("0\n");
    else
    {
        for(int i=x-;i>=;i--) if(col[i]){L=i+;break;}
        for(int i=x+;i<=n;i++) if(col[i]){R=i-;break;}
        if(L== || R==n){printf("INF\n");}
        else printf("%d\n",R-L+);
    }
}

int main()
{
    freopen("explore.in","r",stdin);
    freopen("explore.out","w",stdout);
    n=read();m=read();
    for(int i=;i<=m;i++)
    {
        opt=read();
        if(opt==)
        {
            x=read();y=read();
            solve(x,y,);
        }
        else if(opt==)
        {
            x=read();y=read();
            solve(x,y,);
        }
        else x=read(),find(x);
    }
    return ;
}

70暴力

/*
二分+线段树
By  LXT 林欣彤
*/
#include<complex>
#include<cstdio>
using namespace std;
const int N=1e5+;
int n,m;
int Fog[N<<],lazy[N<<];
int qread()
{
    int x=;
    char ch=getchar();
    while(ch<'' || ch>'')ch=getchar();
    while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
    return x;
}
void PushUp(int rt)
{
    Fog[rt]=(Fog[rt<<]==) || (Fog[rt<<|]==);
}
void PushDown(int rt)
{
    if(lazy[rt])
    {
        Fog[rt<<]=Fog[rt<<|]=lazy[rt];
        lazy[rt<<]=lazy[rt<<|]=lazy[rt];
        lazy[rt]=;
    }
}
void Modify(int l,int r,int rt,int nowl,int nowr,int v)
{
    if(nowl<=l && r<=nowr)
    {
        Fog[rt]=lazy[rt]=v;
        return;
    }
    int mid=l+r>>;
    PushDown(rt);
    if(nowl<=mid)Modify(l,mid,rt<<,nowl,nowr,v);
    if(mid<nowr)Modify(mid+,r,rt<<|,nowl,nowr,v);
    PushUp(rt);
}
int Query(int l,int r,int rt,int nowl,int nowr)
{
    if(nowl<=l && r<=nowr)return Fog[rt];
    int mid=l+r>>;
    PushDown(rt);
    if(nowl<=mid)if(Query(l,mid,rt<<,nowl,nowr)==)return ;
    if(mid<nowr)if(Query(mid+,r,rt<<|,nowl,nowr)==)return ;
    return ;
}
int main()
{
    freopen("explore.in","r",stdin);
    freopen("explore.out","w",stdout);
    scanf("%d%d",&n,&m);
    int p,l,r;
    while(m--)
    {
        p=qread();l=qread();
        if(p==)
        {
            if(Query(,n,,l,l)==){puts("");continue;}
            int L=,R=l,res1=l,res2=l;
            while(L<=R)
            {
                int mid=L+R>>;
                if(Query(,n,,mid,l)==)
                    res1=mid,R=mid-;
                else L=mid+;
            }
            L=l;R=n;
            while(L<=R)
            {
                int mid=L+R>>;
                if(Query(,n,,l,mid)==)
                    res2=mid,L=mid+;
                else R=mid-;
            }
            if(res1== || res2==n)puts("INF");
            else printf("%d\n",res2-res1+);
        }
        else
        {
            r=qread();
            Modify(,n,,l,r,p);
        }
    }
    fclose(stdin);fclose(stdout);
    return ;
}

#include<complex>
#include<cstdio>
#include<iostream>

using namespace std;
const int N=1e5+;
struct node{
    int u,v,w,nxt;
}e[N<<];

int n,m,Enum,tim,tot;
long long ans=1ll<<;
int front[N],in[N],b[N],c[N],fat[N],Log[N],Smin[N][];
bool a[N];

int qread()
{
    int x=;
    char ch=getchar();
    while(ch<'' || ch>'')ch=getchar();
    while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
    return x;
}

void Insert(int u,int v,int w)
{
    e[++Enum].u=u;
    e[Enum].v=v;
    e[Enum].w=w;
    e[Enum].nxt=front[u];
    front[u]=Enum;
}

int find(int x)
{
    if(fat[x]!=x)fat[x]=find(fat[x]);
    return fat[x];
}

void dfs1(int num)
{
    if(num==n-)
    {
        long long res=;
        for(int i=;i<n;i++)
            res+=b[i]*e[i<<].w;
        for(int i=;i<=n;i++)
            fat[i]=i;
        for(int i=;i<n;i++)
            if(!b[i])
            {
                int r1=find(e[i<<].u),r2=find(e[i<<].v);
                fat[r2]=r1;
            }
        for(int i=;i<=m;i++)
            for(int j=i+;j<=m;j++)
                if(find(c[i])==find(c[j]))return;
        ans=min(ans,res);
        return;
    }
    b[num+]=;dfs1(num+);
    b[num+]=;dfs1(num+);
}

void Solve1()
{
    dfs1();
    cout<<ans<<endl;
}

void dfs(int x,int from,int w)
{
    Smin[++tim][]=w;
    if(a[x])b[++tot]=tim;
    for(int i=front[x];i;i=e[i].nxt)
    {
        int v=e[i].v;
        if(v==from)continue;
        dfs(v,x,e[i].w);
    }
}

int Query(int l,int r)
{
    int k=Log[r-l+];
    return min(Smin[l][k],Smin[r-(<<k)+][k]);
}

void ST()
{
    for(int i=;i<=tim;i++)
        Log[i]=Log[i>>]+;
    for(int j=;j<=Log[tim];j++) for(int i=tim-(<<j-);i;i--)
    Smin[i][j]=min(Smin[i][j-],Smin[i+(<<j-)][j-]);
}

void Solve2()
{
    for(int i=;i<=n;i++)
        if(in[i]==)
        {
            dfs(i,,0x3f3f3f3f);
            break;
        }
    ST();
    long long res=;
    for(int i=;i<=tot;i++)
        res+=Query(b[i-]+,b[i]);
    cout<<res<<endl;
}

int main()
{
//    freopen("apple.in","r",stdin);
//    freopen("apple.out","w",stdout);
    scanf("%d%d",&n,&m);
    int u,v,w;
    for(int i=;i<n;i++)
    {
        u=qread()+;v=qread()+;w=qread();
        Insert(u,v,w);Insert(v,u,w);
        in[u]++;in[v]++;
    }
    for(int i=;i<=m;i++)
      c[i]=qread()+,a[c[i]]=;
    int k=;
    for(int i=;i<=n;i++) if(in[i]==)k++;
    if(k==n-)Solve2();
    else Solve1();
    fclose(stdin);fclose(stdout);
    return ;
}

60暴力

/*
形 DP。首先任取一个有苹果的结点为根，构建整棵树，接下来自底向上进行动态
规划。设 f[u]表示以 u 为根的子树中苹果互不连通，且子树中的苹果不与树外的苹果连通的
最小代价。g[u]表示以 u 为根的子树中苹果互不连通所需的最小代价
那么当 u 上有苹果时，
g[u] = sum (f[v]) （v 为 u 的所有子节点） ；
f[u] = g[u] + pre[u] (pre[u]为 u 与其父节点连边的权值） 。
当 u 上无苹果时，
设 s[u] = sum (f[v]) （v 为 u 的所有子节点）
则 g[u] = min (s[u] - f[v] + g[v])
f[u] = min(s[u], g[u] + pre[u])
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>

#define ll long long
#define M 200010

using namespace std;
const ll inf = 10000000000000000ll;
ll f[M], g[M], p[M];
ll nxt[M], head[M], ver[M], to[M], cnt, n, k, sz[M];
bool is[M];
int sta[M];

int read()
{
    int nm = , f = ;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -;
    for(; isdigit(c); c = getchar()) nm = nm *  + c - '';
    return nm * f;
}

void push(int vi, int vj, int v)
{
    cnt++, nxt[cnt] = head[vi], head[vi] = cnt, to[cnt] = vj, ver[cnt] = v;
}

void dfs1(int now, int fa)
{
    if(is[now]) sz[now] = ;
    for(int i = head[now]; i; i = nxt[i])
    {
        int vj = to[i];
        if(vj == fa) continue;
        dfs1(vj, now);
        sz[now] += sz[vj];
    }
}

void dfs2(int now, int fa)
{
    if(sz[now]) f[now] = g[now] = inf;
    ll tot = ;
    for(int i = head[now]; i; i = nxt[i])
    {
        int vj = to[i];
        if(vj == fa) continue;
        if(sz[vj])
        {
            dfs2(vj, now);
            tot += g[vj];
        }
    }
    ll tot2 = , maxx = ;
    for(int i = head[now]; i; i = nxt[i])
    {
        int vj = to[i];
        if(vj == fa || !sz[vj]) continue;
        ll op = min(ver[i], f[vj] - g[vj]);
        tot2 += op;
        maxx = max(maxx, op);
    }
    if(is[now]) f[now] = inf, g[now] = tot + tot2;
    else f[now] = tot + tot2, g[now] = tot + tot2 - maxx;
}

int main()
{
    freopen("apple.in", "r", stdin);
    freopen("apple.out", "w", stdout);
    n = read(), k = read();
    for(int i = ; i < n; i++)
    {
        int vi = read(), vj = read(), v = read();
        push(vi, vj, v);
        push(vj, vi, v);
    }
    for(int i = ; i <= k; i++) is[read()] = true;
    dfs1(, );
    dfs2(, );
    cout << min(f[], g[]) << "\n";
    return ;
}