hihoCoder 1228 Mission Impossible 6 2015北京网络赛 大模拟

题目

题目链接：http://hihocoder.com/problemset/problem/1228

题目来源：2015北京网络赛第四题左右，这场很难，跪烂了。

简要题意：模拟一个文本编辑器的功能，具体看题目吧，太长了。

数据范围：$T\leqslant 20;\quad 0 \leqslant |S| \leqslant 10000$

题解

考虑长度只有四次方，然后要卡时间的话大概最后也就六次方，可以直接数组模拟。

实现

基本上分好块一部分一部分写就行了，注意细节吧。

另外注意各种移动的下标顺序问题。

这题就是个大模拟考验实现能力。

代码

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
LL powmod(LL a,LL b, LL MOD) {LL res=1;a%=MOD;for(;b;b>>=1){if(b&1)res=res*a%MOD;a=a*a%MOD;}return res;}
// head
const int N = 1E4+5;
char s[N];
char in[N];
char cop[N];

class Solver {
public :
    int szlim;
    int len;
    int cur;
    int copyst;
    int copylen;

    bool isInsert;
    bool isCopy;

    Solver(int lim) {
        szlim = lim;
        isInsert = true;
        isCopy = false;
        len = cur = copyst = copylen = 0;
    }

    void enterChar(char ch) {
        if (cur >= szlim || (isInsert && len >= szlim)) return;
        if (isInsert) {
            len = min(len+1, szlim);
            for (int i = len-1; i > cur; i--) {
                s[i] = s[i-1];
            }
        } else {
            len = max(len, cur+1);
        }
        s[cur] = ch;
        cur = min(cur+1, len);
    }

    void enterL() {
        if (cur > 0) cur--;
    }

    void enterR() {
        if (cur < len) cur++;
    }

    void enterC() {
        if (isCopy) {
            int l = min(cur, copyst), r = max(cur, copyst);
            copylen = r-l;
            for (int i = l; i < r; i++) {
                cop[i-l] = s[i];
            }
        } else {
            copyst = cur;
        }
        isCopy = !isCopy;
    }

    void enterB() {
        if (cur == 0) return;
        len = max(len-1, 0);
        for (int i = cur-1; i < len; i++) {
            s[i] = s[i+1];
        }
        cur--;
    }

    void enterD() {
        if (cur == len && !isCopy) return;
        if (isCopy) {
            isCopy = false;
            int l = min(cur, copyst), r = max(cur, copyst);
            int dlen = r-l;
            if (dlen == 0) return;
            for (int i = r; i < len; i++) {
                s[i-dlen] = s[i];
            }
            len -= dlen;
            cur = l;
        } else {
            len = max(len-1, 0);
            for (int i = cur; i < len; i++) {
                s[i] = s[i+1];
            }
        }
    }

    void enterV() {
        if (cur >= szlim || (isInsert && len >= szlim)) return;
        if (isInsert) {
            int vlen = min(len+copylen, szlim)-len;
            len = min(len+copylen, szlim);
            for (int i = len-1; i >= cur+vlen; i--) {
                s[i] = s[i-vlen];
            }
            for (int i = 0; i < vlen; i++) {
                s[i+cur] = cop[i];
            }
            cur += vlen;
        } else {
            int vlen = min(copylen, szlim-cur);
            len = max(len, vlen+cur);
            for (int i = 0; i < vlen; i++) {
                s[i+cur] = cop[i];
            }
            cur = min(cur+vlen, len);
        }
    }

    void enterS() {
        isInsert = !isInsert;
    }

    void handleInput(char ch) {
        if (isCopy && (ch != 'L' && ch != 'R' && ch != 'D' && ch != 'C')) {
            isCopy = false;
        }
        if (ch == 'L') enterL();
        else if (ch == 'R') enterR();
        else if (ch == 'C') enterC();
        else if (ch == 'B') enterB();
        else if (ch == 'V') enterV();
        else if (ch == 'S') enterS();
        else if (ch == 'D') enterD();
        else enterChar(ch);
    }

    void output() {
        s[len] = '\0';
        puts(len == 0 ? "NOTHING" : s);
    }
};

int main() {
    int t, n;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%s", &n, in);
        int len = strlen(in);
        Solver solver(n);
        for (int i = 0; i < len; i++) {
            solver.handleInput(in[i]);
        }
        solver.output();
    }
    return 0;
}