Codeforces Round #484 (Div. 2)

时间：2018-05-20 15:26:45 阅读：206 评论：0 收藏：0 [点我收藏+]

标签：its targe pop quota 队列 top empty 人做 XA

A. Row

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

You‘re given a row with $n$

There are no neighbors adjacent to anyone seated.
It‘s impossible to seat one more person without violating the first rule.

The seating is given as a string consisting of zeros and ones ( $0$

Note that the first and last seats are not adjacent (if $n \neq 2$

Input

The first line contains a single integer $n$

The next line contains a string of $n$

Output

Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".

You are allowed to print letters in whatever case you‘d like (uppercase or lowercase).

Examples

Input

Copy

3
101

Output

Copy

Yes

Input

Copy

4
1011

Output

Copy

No

Input

Copy

5
10001

Output

Copy

No

Note

In sample case one the given seating is maximal.

In sample case two the person at chair three has a neighbour to the right.

In sample case three it is possible to seat yet another person into chair three.

题意：给出一个字符串，0表示空位，1表示有人占位，人和人不能坐一起，且给出的序列没有其他人可以坐下了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mem(a,b) memset(a,b,sizeof(a))
#define P pair< int , int >
const int maxn = 100000+10;
const int INF = 0x3f3f3f3f3f;
char a[1005];
int main()
{
    int n;
    scanf("%d",&n);
    scanf("%s",a+1);
    if(n==1)
    {
        if(a[1]==‘0‘)
            printf("No\n");
        else printf("Yes\n");
        return 0;
    }
    if((a[1]==‘0‘&&a[2]==‘0‘)||(a[n-1]==‘0‘&&a[n]==‘0‘))
    {
        printf("No\n");
        return 0;
    }
    int cnt1=0,cnt2=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]==‘0‘)
            cnt1++;
        else cnt1=0;
        if(a[i]==‘1‘)
            cnt2++;
        else cnt2=0;
        if(cnt1>=3||cnt2>=2)
        {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");

}

In the Bus of Characters there are $n$

Initially the bus is empty. On each of $2 n$

an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it;
an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it.

You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take.

Input

The first line contains a single integer $n$

The second line contains the sequence of integers $w_{1}, w_{2}, \dots, w_{n}$

The third line contains a string of length $2 n$

Output

Print $2 n$

Examples

input

Copy

2
3 1
0011

output

Copy

2 1 1 2

input

Copy

6
10 8 9 11 13 5
010010011101

output

Copy

6 6 2 3 3 1 4 4 1 2 5 5

Note

In the first example the first passenger (introvert) chooses the row $2$

题意：给出n个座位的宽度（每个座位可以坐两个人），给出一个01串，0表示内向的人（喜欢一个人，且喜欢宽度小的座位），1表示外向的人（喜欢别人做，且喜欢宽度大的座位），0和1的个数相等且内向的人，都能选到自己喜欢的座位

这里我们给输入的宽度排序，然后每次遇到0，我们就把入优先队列（就是座位的另一半）（优先队列降序排），然后每次1 就取队首就行了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mem(a,b) memset(a,b,sizeof(a))
#define P pair< int , int >
const int maxn = 200000+10;
const int INF = 0x3f3f3f3f3f;
struct node
{
    int w;
    int id;
    bool friend operator < (const node &u,const node &v)
    {
        return u.w<v.w;
    }
} p[maxn];
bool cmp(const node &u,const node &v)
{
    return u.w<v.w;
}
int main()
{
    priority_queue<node> q;
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&p[i].w);
        p[i].id=i;
    }
    sort(p+1,p+1+n,cmp);
    char s[maxn*2];
    int t=1;
    scanf("%s",s+1);
    for(int i=1; i<=2*n; i++)
    {
        if(s[i]==‘0‘)
        {
            printf("%d ",p[t].id);
            q.push(p[t++]);
        }
        else
        {
            printf("%d ",q.top().id);
            q.pop();
        }
    }
}

You‘re given a tree with $n$

Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.

Input

The first line contains an integer $n$

The next $n - 1$

It‘s guaranteed that the given edges form a tree.

Output

Output a single integer $k$

Examples

input

Copy

output

Copy

input

Copy

3
1 2
1 3

output

Copy

-1

input

Copy

output

Copy

input

Copy

2
1 2

output

Copy

Note

In the first example you can remove the edge between vertices $1$

In the second example you can‘t remove edges in such a way that all components have even number of vertices, so the answer is $- 1$

如果点数是计数，那么直接输出-1。否则，我们可以从任意位置开始dfs，遍历整棵树，遍历的时候。当子树的节点树为偶数的时候，我们可以将其分割，因为偶数段，分割任然为偶数。如果是奇数，那么我们把它加上它的子树节点，然后继续遍历。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mem(a,b) memset(a,b,sizeof(a))
#define P pair< int , int >
const int maxn = 100000+10;
const int INF = 0x3f3f3f3f3f;
int head[maxn],cnt=0,son[maxn];
int ans=0;
struct node
{
    int to,next;
} p[maxn*200];
void add(int u,int v)
{
    p[++cnt].next=head[u];
    head[u]=cnt;
    p[cnt].to=v;
}
void dfs(int u,int f)
{
    son[u]=1;
    for(int i=head[u]; i; i=p[i].next)
    {
        int v=p[i].to;
        if(v!=f)
        {
            dfs(v,u);
            if(son[v]>1&&son[v]%2==0)
                ans++;
            else son[u]+=son[v];
        }
    }
}
int main()
{
    int n,u,v;
    scanf("%d",&n);
    for(int i=1; i<n; i++)
    {
        scanf("%d %d",&u,&v);
        add(u,v);
        add(v,u);
    }
    if(n&1)
    {
        printf("-1\n");
        return 0;
    }
    dfs(1,0);
    printf("%d\n",ans);
}

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~

Codeforces Round #484 (Div. 2)

标签：its targe pop quota 队列 top empty 人做 XA

原文地址：https://www.cnblogs.com/MengX/p/9063367.html

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