CodeForces 518C - Watto and Mechanism（模拟）

Posted 2020-08-16 TianTengtt

题意：有n（1 <= n <= 10^5）个应用，每屏有k（1 <= k <= 10^5）个应用，现在有m（1 <= m <= 10^5）个操作，每次操作会使用一个应用（使用时需滑到应用所在的屏），使用后此应用与前边的相邻应用交换位置，退出此应用后会回到初始屏。问这m次操作总的滚动屏幕次数。

模拟即可。

 

#include<cstdio>  
#include<cstring>  
#include<cctype>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<deque>  
#include<queue>  
#include<stack>  
#include<list>  
typedef long long ll;  
typedef unsigned long long llu;  
const int MAXN = 100 + 10;  
const int MAXT = 100000 + 10;  
const int INF = 0x7f7f7f7f;  
const double pi = acos(-1.0);  
const double EPS = 1e-6;  
using namespace std;  
  
int n, m, k, a[MAXT], loc[MAXT][2];  
vector<int> g[MAXT];  
  
  
  
int main(){  
    memset(loc, -1, sizeof loc);  
    scanf("%d%d%d", &n, &m, &k);  
    for(int i = 0; i < n; ++i)  scanf("%d", a + i);  
    int lur = 0;  
    for(int i = 0; i < n; ++i){  
        int j;  
        for(j = 0; j < k && i + j < n; ++j){  
            g[lur].push_back(a[i + j]);  
            loc[a[i + j]][0] = lur;  
            loc[a[i + j]][1] = j;  
        }  
        ++lur;  
        i = i + j - 1;  
    }  
    llu ans = 0;  
    while(m--){  
        scanf("%d", &lur);  
        ans += (llu)(loc[lur][0] + 1);  
        if(loc[lur][1] == 0){  
            if(loc[lur][0]){  
                int ll = loc[lur][0];  
                int tmp = g[ll - 1][k - 1];  
                loc[lur][0] = ll - 1;  
                loc[lur][1] = k - 1;  
                loc[tmp][0] = ll;  
                loc[tmp][1] = 0;  
                swap(g[ll - 1][k - 1], g[ll][0]);  
            }  
        }  
        else{  
            int r1 = loc[lur][0];  
            int r2 = loc[lur][1];  
            int tmp = g[r1][r2 - 1];  
            loc[lur][1] = r2 - 1;  
            loc[tmp][1] = r2;  
            swap(g[r1][r2 - 1], g[r1][r2]);  
        }  
    }  
    printf("%I64u\n", ans);  
    return 0;  
}