【Educational Codeforces Round 54 (Rated for Div. 2) 】

前言

$这场edu还算打的比较顺利，好久没打cf的我前期竟然比以前还稳$
$稳地莫名其妙，果然第四题栽住了$
$看到题之后自己脑补了一颗dij跑出来的树（赛后才知道这叫最短路树）$
$好久不敲图论自己上去硬莽，写了一个多小时才艰难通过$
$导致第五题没时间看，可能看了场上也敲不出来$
$rating+12,距离上紫还有一些时日，codeforces太好玩了$

---

A. Diverse Substring

题意

$从一个长度为n的字符串中移除一个字符使整个字符串字典序变得最小$
$1<=n<=2\ast 10^5$

做法

$如果可以让字典序变小，而且保证最小，肯定是优先移除前面的，不然移除后面的字典序还是不变$
$所以就找到第一个移除之后能使字典序变小的字符一处就可以$
$什么样的字符移除后能使字典序变小呢$
$思考一下就是str[i]>str[i+1]的时候，移除str[i]会使字典序变小$

坑点

$注意原字符串一直递增时删掉最后一个最优$

代码

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<set>
#include<vector>
#include <time.h>
#include<string.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 2e5+5;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e=exp(1);
const db PI = acos(-1);
const db ERR = 1e-10;

#define Se second
#define Fi first
#define pb push_back
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl

char str[maxn];
int a[maxn];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int len;
    scanf("%d%s",&len,str);
    for(int i=0;i<len-1;i++)
    {
        if(str[i]>str[i+1])
        {
            for(int j=0;j<i;j++) printf("%c",str[j]);
            for(int j=i+1;j<len;j++) printf("%c",str[j]);
            return 0;
        }
    }
    for(int i=0;i<len-1;i++)
    {
         printf("%c",str[i]);
    }
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

---

B. Divisor Subtraction

题意

$不断执行从n中减去n的最小质因子的过程$
$求这个过程会被执行多少次$
$2<=n<=10^{1}0$

做法

$若n为偶数，毫无疑问每次都-2，次数就是n/2$
$若n为奇数，n一定存在一个质因子，而这个质因子一定是奇数$
$n减去这个质因子之后又变为偶数，再/2+1就是答案$

坑点

$注意n本来就是质数时，直接输出1即可$
$还要注意for循环中的i要开longlong$

代码

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<set>
#include<vector>
#include <time.h>
#include<string.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 2e5+5;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e=exp(1);
const db PI = acos(-1);
const db ERR = 1e-10;

#define Se second
#define Fi first
#define pb push_back
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define maxn 100005
int main()
{
    ll n;
    ll ans=0;
    scanf("%lld",&n);
    if(n%2==0)
    {
        printf("%lld\n",n/2);
        return 0;
    }
    for(ll i=3;i*i<=n;i++)
    {
        if(n%i==0)
        {
            printf("%lld\n",(n-i)/2+1);
            return 0;
        }
    }
    printf("1\n");//n为质数的情况
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

---

C. Meme Problem

题意

$a+b=da\ast b=d$
$求a,b是否存在，若存在则输出$

做法

$初中数学题，不明白为什么被放在c题$

代码

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<set>
#include<vector>
#include <time.h>
#include<string.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 2e5+5;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e=exp(1);
const db PI = acos(-1);
const db ERR = 1e-10;

#define Se second
#define Fi first
#define pb push_back
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define maxn 100005
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double x_1,x_2,d;
        scanf("%lf",&d);
        if(d*d-4*d<0)
        {
            printf("N\n");
            continue;
        }
        printf("Y\n");
        x_1=-0.5*(-d+sqrt(d*d-4*d));
        x_2=-0.5*(-d-sqrt(d*d-4*d));
        printf("%.10f %.10f\n",x_1,x_2);
    }
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

---

D. Edge Deletion

题意

$在一个n个点m条边的无向图中起点为1，设初始到达第i个点的最短距离为d[i]$
$现在要求在图上删边，使剩下的边不超过k条，并让尽量多的点d[i]与之前相等$

做法

$我们发现dij跑出来的图最终每个点只有一个前驱$
$那么这就是一棵树，在树上从根节点往下保存边不会影响子孙的d$
$所以跑出dij树之后从根往下保留k条边就可以了。$

坑点

$最短路用dij不要用spfa$
$边权和会超过longlong$

代码

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<set>
#include<vector>
#include <time.h>
#include<string.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 3e5+5;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e=exp(1);
const db PI = acos(-1);
const db ERR = 1e-10;

#define Se second
#define Fi first
#define pb push_back
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl

typedef long long ll;
#define maxm 600005
#define maxn 300005
#define inf 0x3f3f3f3f3f3f3f3f
struct P
{
    int to;
    ll cost;
    bool operator < (const P & a) const
    {
        return cost>a.cost;
    }
};
struct node
{
    int to;
    ll val;
    int nxt;
    int id;
    int s;
}edge[maxm];
int head[maxn],tot;//head和tot记得重置，head重置为-1
int n,m;//点数，边数，不要再main里面重新定义
bool vis[maxn];//每次在dij里面初始化为0
ll dis[maxn];//根据题意初始化为inf可能int可能longlong
void addedge(int x,int y,ll val,int id)
{
    edge[tot].to=y;
    edge[tot].val=val;
    edge[tot].nxt=head[x];
    edge[tot].id=id;
    head[x]=tot++;
}
int pre[maxn],prem[maxn];
int vv[maxn];
void Dijkstra(int s)
{
    memset(vis,0,sizeof(vis));
    fill(dis,dis+n+2,inf);
    dis[s]=0;
    priority_queue<P>q;
    q.push(P{s,0});
    while(!q.empty())
    {
        P p1=q.top();q.pop();
        int u=p1.to;
        if(vis[u])continue;
        vis[u]=1;
        for(int i=head[u];i+1;i=edge[i].nxt)
        {
            int v=edge[i].to;
            if(vis[v])continue;
            if(dis[v]>dis[u]+edge[i].val)
            {
                pre[v]=u;
                prem[v]=edge[i].id;
                dis[v]=dis[u]+edge[i].val;
                q.push(P{v,dis[v]});
            }
        }
    }
}
vector<pii> G[maxn];
vector<int> ans;
int ind[maxn];
int cc,k;
void dfs(int st)
{
    if(cc==k) return ;
    for(int i=0;i<G[st].size();i++)
    {
        cc++;
        ans.push_back(G[st][i].Se);
        if(cc==k) return;
        dfs(G[st][i].Fi);
        if(cc==k) return ;
    }
}
int main()
{
   int u,v;
   ll w;
   memset(head,-1,sizeof(head));
   scanf("%d%d%d",&n,&m,&k);
   for(int i=1;i<=m;i++)
   {
        scanf("%d%d%lld",&u,&v,&w);
        addedge(u,v,w,i);
        addedge(v,u,w,i);
   }
   Dijkstra(1);
   if(k>=n-1)
   {
       printf("%d\n",n-1);
       for(int i=2;i<=n;i++)
       {
           printf("%d ",prem[i]);
       }
   }
   else
   {
       int re=n-1;
       for(int i=2;i<=n;i++)
       {
          G[pre[i]].push_back(pii(i,prem[i]));
       }
        cc=0;
        dfs(1);
        printf("%d\n",k);
        for(int i=0;i<ans.size();i++)
        {
            printf("%d ",ans[i]);
        }
   }
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

---

E. Vasya and a Tree

题意

$给你一颗n个点的树，每个点的权值最初为0$
$有m个操作，每次操作有三个变量v,d,x$
$操作为在v的距离v<=d的子树内所有节点权值+x$
$最终统计树上每个点的权值$

做法

$首先要明确两件事情$
$性质1.每个人的操作只会影响到他的子孙(包括自己)$
$性质2.每个人只会被他祖先的操作所影响(包括自己)$
$也就是说，如果我们能在访问到某个节点时，统计出所有影响到该节点的祖先操作$
$就可以统计出这个节点的最终权值$
$而对于每个操作，我们只要用一个dep数组保存每个深度被增加的值$
$所有深度大于当前节点的操作都会影响到当前节点，如果用线段树就是一个区间求和问题$
$为了减少代码量我们用树状数组，更新时只在本次操作的最深的深度更新$
$这样求一个1-maxdep的前缀和就是所有更新了根节点的操作$
$在求一个1-(nowdep-1)的前缀和就是所有不包含当前节点的操作$
$两个前缀和相减就是当前节点被更新的值$
$为了保证每个操作只影响自己子树内的节点，在dfs退出子树时$
$要将当前根节点的所有修改值还原$

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 3e5+10;
int n,m;
ll tree[maxn],ans[maxn];
vector<int> G[maxn],D[maxn],X[maxn];
void add(int x,int val)
{
    while(x<=n)
    {
        tree[x]+=val;
        x=x+(x&-x);
    }
}
ll sum(int x)
{
    ll ans=0;
    while(x)
    {
        ans+=tree[x];
        x=x-(x&-x);
    }
    return ans;
}
void dfs(int x,int fa,int dep)
{
    for(int i=0;i<D[x].size();i++)
    {
        add(min(D[x][i]+dep,n),X[x][i]);//进子树之前更新
    }
    ans[x]=sum(n)-sum(dep-1);//树状数组变区间查询为两个前缀和相减
    //由于性质2，所以在这个地方就可以直接算出当前节点的最终答案
    for(int i=0;i<G[x].size();i++)
    {
        if(G[x][i]==fa) continue;
        dfs(G[x][i],x,dep+1);
    }
    for(int i=0;i<D[x].size();i++)
    {
        add(min(D[x][i]+dep,n),-X[x][i]);//出子树之后还原
    }
}
int main()
{
    int x,y,z;
    scanf("%d",&n);
    for(int i=1;i<=n-1;i++)
    {
        scanf("%d%d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        D[x].push_back(y);
        X[x].push_back(z);
    }
    dfs(1,0,1);
    for(int i=1;i<=n;i++)
    {
        printf("%lld ",ans[i]);
    }
    return 0;
}