Codeforces Round #408 (Div. 2)(A+B)模拟

A. Buying A House
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.

The girl lives in house m of a village. There are n houses in that village, lining in a straight line from left to right: house 1, house 2, …, house n. The village is also well-structured: house i and house i + 1 (1 ≤ i < n) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.

You will be given n integers a1, a2, …, an that denote the availability and the prices of the houses. If house i is occupied, and therefore cannot be bought, then ai equals 0. Otherwise, house i can be bought, and ai represents the money required to buy it, in dollars.

As Zane has only k dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush’s house to some house he can afford, to help him succeed in his love.

Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 100, 1 ≤ m ≤ n, 1 ≤ k ≤ 100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.

The second line contains n integers a1, a2, …, an (0 ≤ ai ≤ 100) — denoting the availability and the prices of the houses.

It is guaranteed that am = 0 and that it is possible to purchase some house with no more than k dollars.

Output
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.

Examples
input
5 1 20
0 27 32 21 19
output
40
input
7 3 50
62 0 0 0 99 33 22
output
30
input
10 5 100
1 0 1 0 0 0 0 0 1 1
output
20
Note
In the first sample, with k = 20 dollars, Zane can buy only house 5. The distance from house m = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.

In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house m = 3 and house 6 are only 30 meters away, while house m = 3 and house 7 are 40 meters away.
题意：给出n个房子的价格，妹子所在的房子的编号m。和你有的金钱k。
你离妹子的最近距离是多少（编号差*10）.不可以买价格为0的房子。
题解：模拟
代码：

#include<bits/stdc++.h>using namespace std;int n,m,k,x;int main(){    cin>>n>>m>>k;    int ans=1e7;    for(int i=1;i<=n;i++)    {        cin>>x;        if(x&&x<=k)            ans=min(ans,abs(i-m));    }    cout<<ans*10<<endl;}

B. Find The Bone
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Zane the wizard is going to perform a magic show shuffling the cups.

There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.

The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.

Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.

Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.

Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.

The second line contains m distinct integers h1, h2, …, hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.

Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.

Output
Print one integer — the final position along the x-axis of the bone.

Examples
input
7 3 4
3 4 6
1 2
2 5
5 7
7 1
output
1
input
5 1 2
2
1 2
2 4
output
2
Note
In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.

In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.
题意：给出n个杯子，其中m个底下有洞，k次交换。刚开始骨头在1号位置，问最后木头会掉在哪里。
题解：模拟
代码：

#include<bits/stdc++.h>using namespace std;const int N=1e6+10;int vis[N];int n,m,k,x,y;int main(){    scanf("%d%d%d",&n,&m,&k);    for(int i=1;i<=m;i++)scanf("%d",&x),vis[x]=1;    int ans=1;    while(k--)    {        scanf("%d%d",&x,&y);        if(vis[ans]) break;        if(ans==x) ans=y;        else if(ans==y) ans=x;    }    printf("%d\n",ans);}