2019 杭电多校第三场 题解

 

题解

---

 

1001 Azshara's deep sea

Unsolved.

1002 Blow up the city.

 题解：倒着建图，然后将原图出度为零的点用一个新节点连接起来，支配树板题。

参考代码：

#include
using namespace std;
#define LL long long
#define pb push_back
//DAG图q个询问,给你u,v:去掉一个点连同和这个点相连的所有边使得u或v不能到达出度为零的点的方案 
const int maxn=2e5+7;
int n,m,deg[maxn],rt,a[maxn],dep[maxn],val[maxn];
int f[maxn][20];
vector<int>E[maxn],RE[maxn];
void Topsort()
{
    queue<int>q;
    rt=n+1;//图可能不连通,建一个新点连接起来 
    for(int i=1;i<=n;i++) if(!deg[i]){q.push(i);E[rt].pb(i);RE[i].pb(rt);}
    int tot=0;
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        a[++tot]=u;
        for(auto &v:E[u]){if((--deg[v])==0) q.push(v);}
    }
}
int LCA(int x,int y)
{
    if(dep[x]>dep[y]) swap(x,y);
    for(int i=19;i>=0;i--)
    {
        if(dep[y]>dep[x]&&dep[f[y][i]]>=dep[x]) 
            y=f[y][i];
    }
    if(x==y) return x;
    for(int i=19;i>=0;i--)
    { 
        if(f[x][i]!=f[y][i]) 
            x=f[x][i],y=f[y][i];
    }
    return f[x][0];
}
void Work()
{
    dep[rt]=1;
    for(int i=1;i<=n;i++)
    {
        int u=a[i],fa=-1;
        for(auto &v:RE[u]) fa=(fa==-1? v:LCA(fa,v));
        dep[u]=dep[fa]+1;
        f[u][0]=fa; 
        for(int i=1;i<=19;i++) f[u][i]=f[f[u][i-1]][i-1];
    }
}
int main()
{
    int tt; 
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d%d",&n,&m);
        memset(f,0,sizeof(f)); 
        for(int i=1;i<=n+1;i++) 
        {
            E[i].clear(); RE[i].clear();
            dep[i]=deg[i]=0;
        }
        while(m--) 
        {
            int u, v; 
            scanf("%d%d", &u, &v);
            E[v].pb(u); RE[u].pb(v); deg[u]++;
        }
        
        Topsort(); Work();
        int q; 
        scanf("%d", &q);
        while(q--)
        {
            int u, v; 
            scanf("%d%d",&u,&v);
            int lca=LCA(u,v);
            printf("%d\n",dep[u]+dep[v]-dep[lca]-1);
        }
    }
    
    return 0;
}

View Code

1003 Yukikaze and Demons

Unsolved.

1004 Distribution of books

https://blog.csdn.net/liufengwei1/article/details/97703075

#include
using namespace std;
#define maxl 200010


int n,tot,cnt,k;
long long up,dow,ans;
int a[maxl],b[maxl],dp[maxl],h[maxl],c[maxl];
long long sum[maxl];
long long num[maxl];


inline void prework()
{
    scanf("%d%d",&n,&k);
    up=0;dow=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sum[i]=sum[i-1]+a[i],num[i]=sum[i];
        if(a[i]>0) up+=a[i];
        else dow+=a[i];
    }
    num[n+1]=0;
    sort(num+1,num+2+n);
    tot=unique(num+1,num+2+n)-num-1;
    for(int i=0;i<=n;i++)
        b[i]=lower_bound(num+1,num+1+tot,sum[i])-num;
}

inline void upd(int i,int x) 
{
    int t;
    while(i<=tot && i)
    {
        h[i]=c[i];
        for(int k=1;k<(i&-i);k<<=1)
            h[i]=max(h[i],h[i-k]);
        i+=i&-i;
    }
}
inline int qry(int l,int r)
{
    int ret=-n-1,len=r-l+1;
    while(len && r)
    {
        if(len<(r&-r))
        {
            ret=max(ret,c[r]);
            r--;len--;
        }else
        {
            ret=max(ret,h[r]);
            len-=(r&-r);
            r-=r&-r;                
        }
    }
    return ret;
}

inline bool jug(long long mid)
{
    for(int i=1;i<=n;i++)
        dp[i]=0;
    for(int i=1;i<=tot;i++)
        h[i]=-n-1,c[i]=-n-1;
    int l=1,r=n,id,tmp;
    c[b[0]]=0;upd(b[0],c[b[0]]);
    for(int i=1;i<=n;i++)
    {
        id=lower_bound(num+1,num+1+tot,sum[i]-mid)-num;
        if(id>tot)
            continue;
        tmp=qry(id,tot);
        if(tmp<=-n-1)
            continue;
        dp[i]=max(dp[i],qry(id,tot)+1);
        c[b[i]]=max(c[b[i]],dp[i]);
        upd(b[i],c[b[i]]);
        if(dp[i]>=k)
            return true;
    }
        return false;
}

inline void mainwork()
{
    long long l=dow,r=up,mid;
    while(l+1<r)
    {
        mid=(l+r)>>1;
        if(jug(mid))
            r=mid;
        else
            l=mid;
    }
    if(jug(l))
        ans=l;
    else
        ans=l+1;
}

inline void print()
{
    printf("%lld\n",ans);
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        prework();
        mainwork();
        print();
    }
    return 0;
}

View Code

 

1005 Easy Math Problem

https://blog.csdn.net/baiyifeifei/article/details/97798086

#include
using namespace std;
#define int long long
const int size=2e5+5;
const int mod=1e9+7;
typedef long long LL;
bool prime[size]; int p[size];
int tot=0;
int inv6,inv2;
int phi[size];
int sumiiphi[size];
int quick_pow(int a,int b)
{
    register int ans=1;
    while(b)
    {
        if(b&1) ans=1LL*ans*a%mod;
        a=1LL*a*a%mod;
        b>>=1;
    }
    return ans;
}
int inv[205];
void init()
{
    tot=0;
    phi[1]=1;
    for(int i=1;itrue;
    for(int i=2;i)
    {
        if(prime[i])
            p[++tot]=i,phi[i]=i-1;
        for(int j=1;j<=tot&&p[j]*i) 
        {
            prime[i*p[j]]=false;
            if(i%p[j]==0) 
            {
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
            else
            phi[i*p[j]]=phi[i]*phi[p[j]];                
        }
    }
    sumiiphi[0]=0;
    for(int i=1;i1]+1LL*i*i%mod*phi[i]%mod)%mod;
    inv6=quick_pow(6,mod-2);
    inv2=quick_pow(2,mod-2);
    inv[0]=inv[1]=1;
    for(int i=2;i<=200;i++)
    inv[i] = (mod - (mod / i)) * inv[mod % i] % mod;
    for (int i = 1; i <= 200; ++i)inv[i] = inv[i] * inv[i - 1] % mod;
}
int tol;
int hk[size<<1];LL num[size<<1];
int pre[size];
int coeff[205];
void pre_lage(int k)
{
    for(int i=1;i<=k+2;i++) coeff[i]=quick_pow(i,k);
    coeff[0]=0;
    for(int i=1;i<=k+2;i++) coeff[i]=(coeff[i-1]+coeff[i])%mod;
}
LL suf[205],bef[205];
int lage(int n,int k)
{
    if(n<=k+2) return coeff[n];
    bef[0] = suf[0] = 1;
    for (int i = 1; i <= k + 2; ++i) {
        bef[i] = bef[i - 1] * ((n - i) % mod) % mod;
        suf[i] = suf[i - 1] * ((n + i - k - 3) % mod) % mod;
    }
    LL res = 0;
    for (int i = 1; i <= k + 2; ++i) {
        LL s = coeff[i] * bef[i - 1] % mod * suf[k - i + 2] % mod * inv[i - 1] % mod * inv[k + 2 - i] % mod;
        if ((k + 2 - i) & 1)s = -s;
        res += s;
        res = (res % mod + mod) % mod;
    }
    return res;
}
int s;LL n;
int id1[size],id2[size];
inline int ID(int x)
{
    if(x<=s) return id1[x];
    return id2[n/x];
}
void get_h(int k)
{
    s=sqrt(n);
    while(1LL*s*s<=n) s++;
    while(1LL*s*s>n) s--;
    pre[0]=0;
    for(register int i=1;p[i]<=s;i++)
    {
        pre[i]=(1LL*pre[i-1]+quick_pow(p[i],k+1))%mod;
    }
    tol=0;
    for(register LL l=1,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        num[++tol]=r;
        if(r<=s) id1[r]=tol;
        else id2[n/r]=tol;
    }
    pre_lage(k+1);
    for(register int i=1;i<=tol;i++) hk[i]=(lage(num[i]%mod,k+1)-1+mod)%mod;
    hk[0]=0;
    int x=1;
    for(int j=1;j<=tot&&p[j]<=s;j++)
    {
        while(num[x];
        for(int i=tol;i>=x;i--)
        {
            hk[i]=((hk[i]-1LL*(pre[j]-pre[j-1]+mod)%mod*(hk[ID(num[i]/p[j])]-pre[j-1]))%mod+mod)%mod;
        }
    }
}
unordered_mapint> mp;
inline int sum2(int n)
{
    return 1LL*n*(n+1)%mod*(2*n+1)%mod*inv6%mod;
}
inline int sum3(int n)
{
    int ans=1LL*n*(n+1)%mod*inv2%mod;
    return 1LL*ans*ans%mod;
}
inline int S(LL n)
{
    if(nreturn sumiiphi[n];
    if(mp.count(n)) return mp[n];
    LL ans=sum3(n%mod);//n取模 
    for(LL l=2,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        ans=((ans-1LL*S(n/r)*(sum2(r%mod)-sum2((l-1)%mod)))%mod+mod)%mod;
    }
    return mp[n]=ans;
}
int32_t main()
{
    int t;
    scanf("%lld",&t);
    int k;
    init();
    while(t--)
    {
        scanf("%lld%lld",&n,&k);
        mp.clear();
        get_h(k);
        int ans=0;
        for(int i=1;i<=tol;i++)
        {
            ans=(ans+1LL*(hk[i]-hk[i-1])%mod*S(n/num[i])%mod+mod)%mod;
        }
        printf("%lld\n",ans);
    }
}

View Code

 

1006 Fansblog

#include
using namespace std;
typedef long long LL;
unordered_map mp;
LL quick_pow(LL a,LL b,LL mod)
{
    LL ans=1;
    while(b)
    {
        if(b&1) ans=((__int128)ans)*a%mod;
        a=((__int128)a)*a%mod;
        b>>=1;
    }
    return ans;
}
long long factor[110], cnt;
long long Mul_Mod(long long a, long long b, long long c) {
    if (b == 0)
        return 0;
    long long ans = Mul_Mod(a, b / 2, c);
    ans = (ans * 2) % c;
    if (b % 2)
        ans = (ans + a) % c;
    return ans;
}
long long Pow_Mod(long long a, long long b, long long c) {
    if (b == 0)
        return 1;
    long long x = Pow_Mod(a, b / 2, c);
    if (x == 0)
        return 0;
    long long y = Mul_Mod(x, x, c);
    if (y == 1 && x != 1 && x != c - 1)
        return 0;
    if (b % 2)
        y = Mul_Mod(y, a, c);
    return y;
}
bool Miller_rabin(long long n, int timenum = 10) {
    if (n < 2)
        return false;
    if (n == 2)
        return true;
    while (timenum--) {
        if (Pow_Mod(rand() % (n - 2) + 2, n - 1, n) != 1)
            return false;
    }
    return true;
}
long long Gcd(long long a, long long b) {
    long long t;
    while (b) {
        t = a;
        a = b;
        b = t % b;
    }
    return a;
}
void Pollard(long long n);
 
void Factor(long long n) {
    long long d = 2;
    while (true) {
        if (n % d == 0) {
            Pollard(d);
            Pollard(n / d);
            return;
        }
        d++;
    }
}
void Pollard(long long n) {
    if (n == 1)
        return;
    if (Miller_rabin(n)) {
        factor[cnt++] = n;
        return;
    }
    long long i = 0, k = 2, x, y, d;
    x = y = rand() % (n - 1) + 1;
    while (i++ < 123456) {
        x = (Mul_Mod(x, x, n) + n - 1) % n;
        d = Gcd((y - x + n) % n, n);
        if (d != 1) {
            Pollard(d);
            Pollard(n / d);
            return;
        }
        if (i == k) {
            y = x;
            k *= 2;
        }
    }
    Factor(n);
}
int main()
{
    int t;
    scanf("%d",&t);
    LL p;
    while(t--)
    {
        scanf("%lld",&p);
        if(mp.count(p))
        {
            printf("%lld\n",mp[p]);
            continue;
        }
        LL ans=p-1;
        if(Miller_rabin(p-2))
        {
            LL ret=(LL)(((__int128)(p-1))*quick_pow(p-1,p-2,p)%p);
            mp[p]=ret;
            printf("%lld\n",ret);
            continue;
        }
        for(LL i=p-4;i>=998244353;i-=2)
        {
            ans=(__int128)ans*(i+1)%p*(i+2)%p;
            if(Miller_rabin(i))
            {
                LL ret=(LL)(((__int128)(p-1))*quick_pow(ans,p-2,p)%p);
                mp[p]=ret;
                printf("%lld\n",ret);
                break;
            }
        }
    }
}

View Code

1007 Find the answer

#include
#define maxl 200010
using namespace std;

int n,m;
int a[maxl],dy[maxl],ans[maxl];
struct anode
{
    int id,val;
}aa[maxl];
struct node
{
    int l,r;
    long long sum;
}tree[maxl*4];
int b[maxl];

inline bool cmp(const anode &x,const anode &y)
{
    return x.val<y.val;
}

inline void add(int i,int x)
{
    while(i<=n)
    {
        b[i]+=x;
        i+=i&-i;
    }
}

inline int sum(int i)
{
    int ret=0;
    while(i)
    {
        ret+=b[i];
        i-=i&-i;
    }
    return ret;
}

inline void build(int k,int l,int r)
{
    tree[k].l=l;tree[k].r=r;
    if(l==r)
    {
        tree[k].sum=aa[l].val;
        return;
    }
    int mid=(l+r)>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;
}

inline void prework()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),aa[i].id=i,aa[i].val=a[i];
    sort(aa+1,aa+1+n,cmp);
    for(int i=1;i<=n;i++)
    {    
        dy[aa[i].id]=i;
        b[i]=0;
    }
    for(int i=1;i<=n;i++)
        add(i,1);
    build(1,1,n);
}

inline void upd(int k,int l)
{
    if(tree[k].l==tree[k].r)
    {
        tree[k].sum=0;
        return;
    }
    int mid=(tree[k].l+tree[k].r)>>1;
    if(l<=mid)
        upd(k<<1,l);
    else
        upd(k<<1|1,l);
    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;
}

inline int query(int k,long long res)
{
    if(tree[k].l==tree[k].r)
    {
        if(tree[k].sum>res)
            return tree[k].l-1;
        else
            return tree[k].l;
    }
    if(tree[k<<1].sum<res)
        return query(k<<1|1,res-tree[k<<1].sum);
    else
        return query(k<<1,res);
}

inline void mainwork()
{
    int l;
    for(int i=n;i>=1;i--)
    {
        upd(1,dy[i]);
        add(dy[i],-1);
        l=query(1,m-a[i]);
        ans[i]=(i-1)-sum(l);
    }
}

inline void print()
{
    for(int i=1;i<=n;i++)
        printf("%d ",ans[i]);
    puts("");
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        prework();
        mainwork();
        print();
    }
    return 0;
}

View Code

1008 Game

Unsolved.

1009 K Subsequence

https://blog.csdn.net/baiyifeifei/article/details/97706037

#include
using namespace std;
typedef pair<int, int> pii;
const int maxn = 1e4;
const int inf = 0x3f3f3f3f;
struct edge {
    int to, cap, cost, rev;
    edge() {}
    edge(int to, int _cap, int _cost, int _rev) :to(to), cap(_cap), cost(_cost), rev(_rev) {}
};
int V, H[maxn + 5], dis[maxn + 5], PreV[maxn + 5], PreE[maxn + 5];
vector G[maxn + 5];
void init(int n) {
    V = n;
    for (int i = 0; i <= V; ++i)G[i].clear();
}
void AddEdge(int from, int to, int cap, int cost) {
    G[from].push_back(edge(to, cap, cost, G[to].size()));
    G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
}
int Min_cost_max_flow(int s, int t, int f, int& flow) {
    int res = 0; fill(H, H + 1 + V, 0);
    while (f) {
        priority_queue int, int>, vectorint, int>>, greaterint, int>> > q;
        fill(dis, dis + 1 + V, inf);
        dis[s] = 0; q.push(pair<int, int>(0, s));
        while (!q.empty()) {
            pair<int, int> now = q.top(); q.pop();
            int v = now.second;
            if (dis[v] < now.first)continue;
            for (int i = 0; i < G[v].size(); ++i) {
                edge& e = G[v][i];
                if (e.cap > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to]) {
                    dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
                    PreV[e.to] = v;
                    PreE[e.to] = i;
                    q.push(pair<int, int>(dis[e.to], e.to));
                }
            }
        }
        if (dis[t] == inf)break;
        for (int i = 0; i <= V; ++i)H[i] += dis[i];
        int d = f;
        for (int v = t; v != s; v = PreV[v])d = min(d, G[PreV[v]][PreE[v]].cap);
        f -= d; flow += d; res += d*H[t];
        for (int v = t; v != s; v = PreV[v]) {
            edge& e = G[PreV[v]][PreE[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}
int a[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        for(register int i=1;i<=n;++i) scanf("%d",&a[i]);
        int ss=0,s=1,t=2*n+2,tt=2*n+3;
        init(tt+1);
        AddEdge(ss,s,k,0);
        AddEdge(t,tt,k,0);
        for(register int i=1;i<=n;++i)
        {
            AddEdge(s,i+1,1,0);
            AddEdge(i+1+n,t,1,0);
            AddEdge(i+1,i+1+n,1,-a[i]);
            for(register int j=i+1;j<=n;++j)
            {
                if(a[j]>=a[i])
                {
                    AddEdge(1+i+n,1+j,1,0);
                }
            }
        }
        int ans=0;
        printf("%d\n",-Min_cost_max_flow(ss,tt,inf,ans));
    }
    return 0;
}

View Code

1010 Sindar's Art Exhibition

Unsolved.

1011 Squrirrel

题解：题目意思给你一棵树，每条边有一个权值。

让你求：在将一条边的权值变为零的情况下，一个点到其最远叶子节点的距离的最小值为多少。

思路：

先考虑在不使一条边变为零的情况下：我们先dfs处理出，每一个节点到其叶子节点的最远距离l[i]，然后求出其沿着父亲节点的最远距离fa[i],则对于这个节点的就是max(l[i],fa[i])，然后在所有节点中的最大距离取个最小值就行了。 现在是可以任意一条边变为零，那么我们先dfs，记录每个节点到其叶子节点的最大值，次大值，第三大值，然后求出其沿着父亲节点的最大值，对于将边变为零的情况，我们只要在DP时加上一维，判断是否江一条边变为零就行了。

 

参考代码：

#include
using namespace std;
typedef long long ll;
#define pii pair
#define pil pair
#define PI acos(-1.0)
#define mkp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
const int INF=0x3f3f3f3f;
const ll inf=0x3f3f3f3f3f3f3f3fll;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
const int maxn=2e5+10;
int T,n,cnt,head[maxn];
struct Edge{
    int to,nxt,w;
} edge[maxn<<1];
void AddEdge(int u,int v,int w)
{
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].nxt=head[u];
    head[u]=cnt++;    
}
int dis[maxn],ans[maxn];
int f[maxn],s[maxn],t[maxn],fa[maxn][2];
int fi[maxn][2],se[maxn][2],th[maxn][2];

void Init()
{
    cnt=0; mem(head,-1);
    mem(dis,0); mem(ans,0);
    for(int i=0;i<=n;++i)
    {
        fi[i][0]=fi[i][1]=0;
        se[i][0]=se[i][1]=0;
        th[i][0]=th[i][1]=0;
        fa[i][0]=fa[i][1]=0;
        f[i]=t[i]=s[i]=0;
        dis[i]=0; 
    }    
}

void dfs1(int x,int pre)
{
    for(int i=head[x];~i;i=edge[i].nxt)
    {
        int v=edge[i].to,w=edge[i].w;
        if(v==pre) continue;
        dis[v]=w;
        dfs1(v,x);
        if(w+fi[v][0]>fi[x][0])
        {
            th[x][0]=se[x][0];
            t[x]=s[x];
            se[x][0]=fi[x][0];
            s[x]=f[x];
            fi[x][0]=fi[v][0]+w;
            f[x]=v;
        }
        else if(w+fi[v][0]>se[x][0])
        {
            th[x][0]=se[x][0];
            t[x]=s[x];
            se[x][0]=fi[v][0]+w;
            s[x]=v;
        }
        else if(w+fi[v][0]>th[x][0])
        {
            th[x][0]=fi[v][0]+w;
            t[x]=v;    
        }
    }
    fi[x][1]=min(fi[f[x]][0],dis[f[x]]+max(fi[f[x]][1],se[f[x]][0]));
    se[x][1]=min(fi[s[x]][0],dis[s[x]]+max(fi[s[x]][1],se[s[x]][0]));    
}    

void dfs2(int x,int pre)
{
    ans[x]=min(max(fi[x][0],fa[x][1]),max(max(fi[x][1],se[x][0]),fa[x][0]));
    for(int i=head[x];~i;i=edge[i].nxt)
    {
        int v=edge[i].to,w=edge[i].w;
        if(v==pre) continue;
        
        if(f[x]==v)
        {
            fa[v][0]=max(fa[x][0],se[x][0])+w;
            fa[v][1]=min(max(fa[x][0],se[x][0]),min(max(fa[x][0],max(se[x][1],th[x][0])),max(se[x][0],fa[x][1]))+w);    
        }
        else if(s[x]==v)
        {
            fa[v][0]=max(fi[x][0],fa[x][0])+w;
            fa[v][1]=min(max(fi[x][0],fa[x][0]),min(max(fa[x][0],max(fi[x][1],th[x][0])),max(fi[x][0],fa[x][1]))+w);
        }
        else
        {
            fa[v][0]=max(fi[x][0],fa[x][0])+w;
            fa[v][1]=min(max(fi[x][0],fa[x][0]),min(max(fa[x][0],max(fi[x][1],se[x][0])),max(fi[x][0],fa[x][1]))+w);
        }
        dfs2(v,x);
    }
}

int main()
{
    T=read();
    while(T--)
    {
        n=read();
        Init();
        int u,v,w;
        for(int i=1;ii)    
        {
            u=read();v=read();w=read();
            AddEdge(u,v,w);AddEdge(v,u,w);    
        }
        dfs1(1,0);
        dfs2(1,0);
        int val=INF,pos;
        for(int i=1;i<=n;i++)
        {
               if(ans[i]i;
            else if(ans[i]==val&&ii;
        }
        printf("%d %d\n",pos,val);
    }
    
    return 0;    
}
/*
1
5
1 5 1
1 2 1
2 3 2
3 4 1
*/

View Code

 

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